KING'S COLLEGE SCHOOL, LONDON. ii|j Numerous |Uusi
126
ELEMENTS OF PLANE TRIGONOMETRY. BY JAMES HANN, TOBSIERLY MATHEMATICAL MASTER; KING'S COLLEGE SCHOOL, LONDON. ii|j Numerous |Uusirations. THIRD EDITION. LONDON: YIETUE BROTHERS & CO., 26, IVY LANE, PATERNOSTER ROW. 18G7.
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Transcript of KING'S COLLEGE SCHOOL, LONDON. ii|j Numerous |Uusi
ELEMENTS
OF
PLANE TRIGONOMETRY.
BY
JAMES HANN, TOBSIERLY MATHEMATICAL MASTER; KING'S COLLEGE SCHOOL,
LONDON.
ii|j Numerous |Uusirations.
THIRD EDITION.
LONDON: YIETUE BROTHERS & CO., 26, IVY LANE,
PATERNOSTER ROW. 18G7.
Q.A 5
ScA
632957 PKEFACE TO THE THIKD EDITION.
[n submitting this Rudimentary Treatise on Plane LYigonometry, after its having been carefully revised ind corrected, the author would also respectfully Jivite the reader's attention to the collection of Mathematical Tables, forming part of this Series of Scientific Treatises ; they will be found amply suf- icient for all the practical purposes of Trigonometry, md will therefore furnish all the aid necessary in computations connected with the present subject.
The author has given, to illustrate the principles, a great number of examples fully worked out, and which will be of service to those who have not the aid of a teacher.
"In compiling the work, the best authors, whether French or English, have been consulted. Amongst others, those of Bonnycastle, Cape, De Morgan, Gaskin, Hall, Hind, Hymers, Snowball, Woodhouse, and Gre- gorywith Davies's edition of Button's Course.
The problems have been principally taken from the Ladies and Gentleman's Diaries, the Cambridge Pro¬ blems, and Laybourn's Repository.
"The demonstration ofDemoivre's Theorem is taken trom an able French work on Trigonometry, byLefebure De Fouroy."
J. HANK
fZf y.ltf
CONTENTS.
CHAPTER X. Page
The object of Trigonometry. French and English Measures . . 1 Angular measures 2, 3 Trigonometrical ratios 8 Tracing the various trigonometrical functions 12, 13 Values of sine, cosine, tangent, cotangent, secant and cosecant of
30°, 45°, and 60° 14, 15
cot (A±£)z
'• tan A tan B cot A cot JB+ 1
25, 26
CHAPTER II. sin (A zh B) = sin A cos £ zb. cos A sin B cos (A dc B) = cos A cos B + sin A sin B .
v * "sin SA, cos 3A, &c 27, 28 t,n(A!±£)=^-A±t!iriB ^ V J 1 -+- t 30, 31
cot B ± cot A Inverse trigonometrical functions 32, 33 Values of the sine and cosine of 15°, 18°, 36°, 54°, 74°. . .34, 35
CHAPTER III. Solution of triangles. Values of cos A, cos B, (fee. . . . 51, 52, 53 Values of sin A, sin %A, cos %A, tan %A 55 Area of a triangle 56 Ambiguous fcase 58 Area of a triangle in terms of the radius of the inscribed circle . 59 Area of the triangle in terms of the radius of the circumscribed
circle, <fec. 60 Practical remarks on the solution of triangles . . . . .75
CHAPTER IV. Polygons 80
CHAPTER V. Demoivre's theorem 104
TBIGrON OMETRY.
CHAPTER I.
1. •_ Teigonometet was oiiginally considered to be the doctrine of triangles, but in its present improved state it has a much more extensive signification, which we shall here¬ after shew even in this rudimentary treatise.
2. In estimating angular measures, we suppose the right angle to be the primary one, and to be divided into 90 equal parts, each of which is called a degree; each degree is supposed to be divided into 60 equal parts, each of which is called a minute ; each minute- is supposed to be divided into 60 equal parts, each of which is called a second, and so on to thirds, fourths, &c. Here one degree is considered as the angular unit.
_ 3. Modern French writers, instead of using the -sexa^c- sima! dmsion, use the centesimal; audit is to be regretted that the latter is not universally used, from the great ease with which all calculations are made in that division
"We shall, however, shew how to reduce French into Jinglish measures, and vice vend,
If ^ and .F represent the number of English degrees and .trench grades m the same angle,
E F E F 90 ~ 100 or IT = fo5
... 10^- 7/4-^ 10 10'anCli' 9 ~9'
VUi?0hrCirCTfCvnC6i0f a circle is kn0WI1 to be about
H,n o;f9, r3 ! 0r' iu 0ther words. the ratio of the circumference to the diameter is represented by 3-14159 • tor this number writers generally put the Greek letter tt. '
B
A TRIGONOiTETItY.
circumference = tt D\ where D is the diameter, or 2 it r, where r is the radius of the circle.
Hence the length of the arc of a quadrant is of a semi- Ji
circle, or 180°, is 7rr; and of 270°, or three quadrants, is^-- 2*
"Now if any arc a subtend an angle of A0, then since — 2
subtends 90°, and that by Euclid vi. 33, angles are propor¬ tional to the arcs which subtend them,
Ao:90o::a:—;
tt r Prom this expression any one of the quantities may be
found when the others are given. Ex. 1. Eind the length of an arc of 45° of a circle whose
radius is 10 feet; 450 _1!2! i.
"" IT * 10' 45°
a = -—X 10 X 3-14159 = 7-8539 feet. 180°
5. Most modern writers on Trigonometry take also for the unit of angular measure the number of degrees in an angle, subtended by an arc equal to the radius*. If TJ0 re¬ present that angle, then by equation (1),
* If ACB be an angle at the centre of a circle, subtended by an arc equal to the^ radius of the circle, then, since by the 33rd Proposition of the Cth Book of Euclid, the angles at the centre of a circle are to each other as the arcs on which they stand,
Angle ACB : four right angles :: arc AB : cir¬ cumference, but AB is an arc equal to the radius,
.*. Angle ACB : four right angles :: r : 2 tt r :: 1 : 2 tt,
ACB =f0Ur anglC^ which, being independent of r, is constant for any circle; it may therefore be used to measure other angles.
TEIGONOMETEY. 3
180° r 180° 180° U° = 570,29578. TT T TV 3*14159
Hence, A0 = 570-29578 ox A° = TT° (2).
And since U0 = 570,29578 is constant, ^ a . arc vanes as — , i.e. as —r:— ;
r radius and taking ZP2 as the unit, we have
A0 = — which is called the circular measure of the angle. r
Prom equation (2) we see that the measuring unit, 77°,
must be multiplied by the fraction — to find the angle;
thus if the circular measure of an angle be —> then
A0 = (57°-29578) = 280-64789. g
If the circular measure be —- then 5
6 12 A0 = - (570-29578) = — (570-29578) = 680-754936.
5 10 Now, suppose we take an angle of 22° 27' 39", then this is
put into decimals at once by the centesimal division, without putting down any work on paper, it being 220,2739 ; whereas, by the sexagesimal, we must proceed in the following manner:
60J 39 60; 27-65
22*4608 If we wish to find how many grades and minutes' are
( contained in this angle, here U = 22-4608
' U 2-4956
lU Zi + 24*9564, which at sight is 24* 95' 64".
n 2
4 TBIGONOMETRY.
Find the number of degrees and minutes in 46g 56' 36". F =46-5636 F — = 4*65636 10
E^F =41-90724 10
60 54-4344
60 26-064
E0 = 41° 54' 26". (1) If Fl and F", F' and F'' represent the magnitude of a
French and English minute and second respectively, shew that F' __ 3-32 F" _ 3^ ~f'~ 2^'
F* — 1 French min. ; F! x 100 X 100 = a quadrant,
F' — 1 English min.; F1 X 60 x 90 = a quadrant,
hence, F' X 10000 = F' X 5400, F' x 50 = F' x 27, i^7' 27 3-32
^ ^ 5^ " 2^* Also ,F" = 1 French second;
x 100 X 100 x 100 = a quadrant, J?'' = 1 English second; F" x 60 x 60 x 90 = a quadrant
"hence, F" X 1000000 = F" X 324000, F" X 250 = X 81 ; F" H ^ ^ 250 ^ 2*53
(2) Compare the interior angles of a regular octagon and dodecagon:
360° In a polygon of n sides, A0 = 180° — .
o6oo
In octagon = 180° - — = 180° - 45° = 135°.
^ TKIGONOMETRY. 5
360° In dodecagon = 180° - — = 180° - 30° = 150°,
1 Ji 135 : 150 : : 9 : 10.
The ratio is therefore 9 to 10. (3) The earth being supposed a sphere, of which the
diameter is 7980 miles, find the length of an arc of 1°. 180° a 180° a 180° a
' X r' " — "V" X 7980 ~ir X 3990' 2
180° but = 570-29578;
TT a
1° = 570-29578 x 3990' 3990
57-29578 = 69-6 miles.
(4) Find the diameter of a globe when an arc of 25° of the meridian measures 4 feet.
180° a 180° 4 180° 8 A° = X -: ••• 253 = x - = X —;
jt > k v x 2r
... 2r = — x = 57-29578 x tt 25 25
= 18*3346 feet. Answer. (5) Pind the number of degrees in a circular arc 30 feet
in length, of which the radius is 25 feet. 180° a 180° 30 180° 6
^0= X = x— = X- TT 25
= 570,29578 X - = 680-75493. 5
(6) Pind the number of degrees in an angle of which" the circular measure is *7854, the value of tt being 3*1416.
180° a 180° . , 180° A0 = x- =—■— X*7854 = —— = 45°.
tt r 3*1416 4 (7) The interior angles of a rectilineal figure are in arith¬
metical progression, the least angle is 120°, and the common difference 5° ; required the number of sides.
TKIG ON 0 JIETHY.
Sum of angles = mr — 27r by Euc._(i- 32), or 120°+125°+130°, &c., to n terms = 180°^— 360°,
{240+(n _ 1) 5} ^ = 180» - 360, Ji
120n+5n2~ — = 180» - 360, Ji
240n+5n2 - 5n = 360;i — 720, 5n2 — I25n = — 720,
625 625 Pc 625 - 576 ~ 4 4 ~~
49 T'
25 2~ 2'
n 25 4- 7 ^ „
= = 16 or 9. 2
The last is tlie congruent value of w, since no angle can be so great as 180° ; the figure has nine sides.
(8) One regular polygon has two sides more than another, and each of its angles exceeds each angle of the other poly¬ gon by 15°; find the number of sides in each.
nA= mr — 27r, equation (1), (ft — 2) (A. — 15) = (n- 2 x 2) tt,
nA — 2A — 15ft + 30 = nir — Air, equation (2). Subtracting equation (2) from (1),
2A + 15ft — 30 = 27r, 2 A = 27r — 15ft + 30,
2-17 — 15ft + 30 A = , and from equation (1)
2/ mr—27r 180ft—360
A = ; ft ft
27r — 15ft + 30 180ft-360 2 ~ n '
390ft- 15ft2 = 360ft- 720, 15ft2 - 30ft =*. 720,
V - 2ft = 48, ft2 — 2ft + 1 = 49,
THIGOXOMETEY. 7
n — 1 = zb 7, ix = 1 + 7 = 8 or — G,
and n — 2 = 6. An octagon and a hexagon.
(9) The angles in one regular polygon are twice as many as in another polygon ; and an angle of the former is to an angle of the latter as 3 : 2 ; find the number of sides.
Let n = number of sides in 1st, A — each angle, 2tt = number of sides in 2nd, B = each angle,
B n — 1 , 3. — = also=-: A n — 2 2'
2n — 2 — 3n — 6^ 91 = 4,
2n = 8. (10) The angles of a quadrilateral are in increasing geome¬
trical progression, and the difference between the third angle and the fourth part of the first is 90°; find the angles.
Let A, Ar, Ar2, Ar*, be the angles; A (1 + r + r2 + r3) = 360°,
and A (r2 — = 90°;
... (x + r + r2 + ^3) ^ (4r2_ . ... r3 —3r2 + r + 2 = 0;
.-.t4— 3r3 + r2 + 2r= 0,
(,.i_3ri+^r!) — (r2—-r) ~ 4 v 2 4 4 2 4
"r 2r 2 2 2' f2= 2r; r = 2.
But A (r2— ^ = 90°; .-.^=90°
\bA = 360°; .-.A — 24°. And the angles are 24°, 48°, 96°, 192°.
8 TlilGONOilETEY.
6. TliIG0N03IETIlICAL EaTIOS OR DEFINITIONS.
Let BA 0 "be any angle, and from any point D in AB let fall the perpendicular DP on A C, then if we represent the angle BA G l?y jyp ^ ^ —■ is the sine of A; is the cosine of A ; JLJ-J J) p
is the tangent of A \ is the cosecant of Ay
AD AP is the secant of A; —- is the cotangent of A.
-a. Jl JJJL «■ Por the sake of abbreviation the above quantities are
generally put sin A, cos A, tan A, sec A, cosec A, cot A. 1 — cos A is defined to be the versed sine of A, or vers A.*
By the 47th Prop, of the 1st Book of Euclid, AP2 + DP2 = AD2 (1)
* It may be perhaps necessary to re¬ mark, that the following definitions have been given by most English writers till within the last few years. In the annexed figure, take any arc AB, draw -Z?P and A T each perpendicular to the diameter AE, and produce OB to meet AT in T, then BP is called the sine of the angle A CB to the radius CB; CP is called the cosine; AT the tangent; CTthe secant; AP the versed sine; OT' the cotangent; CT' the cosecant.
If we take the arc Ab greater than one quadrant and less than two, then bp is called the sine; Cp the cosine; AB the tangent; CR the secant; Ap the versed sine.
If th6 arc Abd be greater than two quad- , rants but less than three, then de is called the sine; Ce the cosine; AB the tangent; CB the secant; Ae the versed
If the arc Abdj be greater than three quadrants but less than four, then fg is called the sine; Cg the cosine; Ac the tangent; Cc the secant; and Ag the versed sine.
. Let the angle ACB = A,r = radius CB, then since CP2 + PB^= CB2 by 47th Prop. 1st book of Euclid, we have cos2 A + sin2 A = r2 (I)
Ako AC2 + AT2= CT2, or r2 -f tan2 A = sec2 A (2)
IltlGONOJTETRY. 9
Divide both sides of this equation by AD\ and we have AP- ^ BP^ _ AIP. ZW- ' AD* ~~ AIT-'
(AFs? . /i>-P\2
-4P , but is the cosine of A, and —- is the sine of A, by the -4-Z/
above definitions; cos2 A + sin2 A =1 (2).
Prom this equation we have, cos2 A = 1 — sin2 A; cos A = v^(l — sin2
and sin2 A = 1 — cos2 A, sin A = \/(l — cos3 A) ;
j)j> ^p and since by the definitions —— = sin A, and = cos Ay ^oLJJ -A-UJ we have DP = AD sin A} and AP = AD cos A.
__ . . DP AD sin A sin A = ZP = Zflco7I = cos^'
From (1) cos2 A = r^—sim A, or cos A= sj(r*2 — sin2 A) and sin2 A — r2 — cos2 A, or sin A — s/ (r2 — sin2 A)
From (2) tan2 A — sec2 A — r2. The triangles BOP, TO A, OGT, and oCB, are all similar,
AT BP tan A sin A "AC~CP0T r ^cos A
r sin A or tan A = (3) cos A cot A cos A
r sin A , rcoaA or cot A = —; (4) sin A v '
sec A r r2
= 7 or sec A = (5) r cos A cos A cosec A r T2 fn. =-:—- or cosec A — - (p) r sin A sin A
versing = r — cos A (7) B 5
10 TEIG ON OMETTtY.
cot A = —— = AP AD cos A cos A
sec A
DP AD
AD sin A AD
sin A 1
2 ^ sec2 A = —— =
AP AD cos A AD AD ~DP
AP2 + DP2
cosec A = —- =
cos A1
1 AD sin A sin A1
DP\2
1 + /DPV_ \AP)~
1 + tan2 A • AP2 AP2
or by dividing equation (1) by AP2, /AP\2 (DP\*__ (AD\2
\AP) + \AP) " \AP) 7
or 1 -)- tan2 A = sec2 A; sec A = ^(1 + tan2 A),
Also dividing equation (1) by DP2, we have
cot2 A + 1 = cosec2 A ; cosec A = ^(1 + cot2 A).
(1) The sine of an angle is equal to the cosine of its compkment.
Since L ADP = 90° — A, AP
and sin ADP = —- — cos A, ./LjU
DP cos ADP = -pf = sirv A ;
that is, sin (90° — A) = cos A, and cos (90° — A) — sin A.
(2) To shew that sin A = sin (180° — A). c
Let B0Pl be greater than a right angle: draw perpen¬ dicular to AB; make the angle J30P equal to the angle AOPv , and OP = OPv and let fall the' iTl v<TX perpendicular PjV, then the tri¬ angles Pl OjSr
l and PON are evi¬ dently equal.
) p
N /o
p. Nf
Pa
T HIG ON" OMETET • 11
P N — - . jPiV sin BOP, ==^; sin.Z?OP = sin (180°—BOP^ ;
^1 but P.iV, = PiV, and OP, = OP;
sin iJOP, = sin (180° — -BOP,) : that is, tho sine of an angle is equal to the sine of the sup- plement of that angle :
OJSf, ON cos B OPj OP,
cos BOP — cos (180°
" OP7
. on ■ BOP J — Qp;
cos BOP, = — cos (180° — ^ or the cosine of an angle is equal to minus the cosine of its supplement.
(3) Sin (—A) = —sin A, and cos (—A) = cos
If the line OP revolve round the point 0 from OB upwards till it describes an angle NOP = Ay then if it revolve downwards o from OB till it describes the angle WOP, = NOP, then JSTOPl = —A.
In
PJV sin A ■= —, and sin JVOP, = sin (— A) = ,
^ PjiV PJST OP, _ OP'
sin (—A) = — sin A; OJV OJSf ON
cos A = —, and cos iVOP, = cos (— A) = — = ;
cos A = cos (— A). Prom these we readily see that
tan A = — tan (— A), cot A = — cot (— A), sec A = sec (— A),
cosec A = — cosec (— A). After one revolution is completed, the sines, cgsines, &c.,
take the samq values as before; therefore the sine of any angle is the same as the sine of 360° + that angle, or
sin A =• sin (27r + As ;
12 TlilGONCLUETIiY.
In the same way, sin A = sin (47r +-£)>
and for the general form, where n is a whole number, we have
sin A = sin {2?t7r + A). Similarly,
sin (tt — A) = si-i (."tt — A) = sin (Bit — A) ; or, generally,
sin (tt — A) = sin { 2n -{- 1) tt — A} ; but sin A = sin (tt — A) ;
sin (2«x + A) = sin {(2?i + 1) tt — ^j. Also sin A = —sin {(2;i + 1) tt + -^}
= — sin {2mr — A)j cos A = eos (2n7r + A) == — cos {(2n + 1) tt — A}, tan A = tan (2n7r + A) = — tan {(2n + 1) tt — A], sec A = sec (2?i7r + A) = — sec {{2n + 1) tt — A).
(4) In the figure at page 10, it is clear that if we suppose the line OP, originally coinciding with OB, to revolve round 0 as a centre, in each of the quadrants the various trigonometrical quantities will have precisely the same value; for the sake of distinction these quantities are affected with different signs. Thus, suppose ON = 0iVJ = x, then ON, measured to the right of 0, is called + cc, and OJSFi measured to the left of 0, is called — x. Also, any
-line JPJV above the line AB is considered positive, and any line below the same line is considered negative; thus if iW = + y, then is—y. This is purely cohven- tional, for we might have taken lines to the left to be positive, and those to the right to be negative, and so on; but when we once fix on the positive direction, the negative direction must necessarily be opposite to it.
If we suppose the line OP to revolve upwards, the angle PON is considered positive; but, if downwards, the angle PzO]Sr is considered negative : thus, the angle PON is + A, and the angle PzON is — A.
(5) We may now proceed to trace the values of the srne, cosine, (fee., throughout the four quadrants. In the first
PN ... . ON quadrant which is the sine of A, is positive ; and
which is the cosine of A} is also positive ; it is clear that as
TMGONOMETRY. 13
the sine increases, the cosina decreases, and when OP coin¬ cides with OC\ then Pj^=OP9 and OJSF—O) hence the sin 90° =1 and cos 90° = 0. In the second quadrant the
om sine is also positive, but the cosine or —— is negative,
.1
since OiVi is measured in the opposite direction to OJSf; as OP revolves from 00 towards OA the sine decreases, and the cosine increases negatively; so that when OP coincides with OA the sine is 0 and the cosine is —1, that is, sin 180° = 0 and cos 180° = — 1.
In the third quadrant both the sine and cosine are negative, and when OP coincides with OD, then the sin 270° = -1, and the cos 270° = 0. In the fourth quadrant the sine is negative but the cosine positive, and when the line OP has completed a whple revolution by coinciding again with OB, then we have the
sine 360° = 0, and cosine 360° = 1.
Since tan A = ^ ^ ; in the first quadrant, both sin A cos A 1 ^ and cos A being positive, tan A is also positive, and when
sin 90° 1 A = 90° we have tan 90° = —^— = - = infinity; that is,
the tangent of 90° is infinite.
In the second quadrant tan A = an^ is there- 1 — cos A „ .. _ . „ ^ sin 180° 0 lore negative, and tan 180° = - Q = —r = 0. COS loll ■ J.
In the third quadrant tan A = —5 > and is thereforo COS J X sil1 2700 - 1 - * positive, and tan 270° = = ——= — infinity. ' cos 270° 0 J
In the fourth quadrant tan A = ^ A, and eonse- —p COS Jx
quently negative, and tan 360° = ^ ggQo = j =
(6) To find sine, cosine, &c., of 30°, 60°, 45°. Let the angle be 60°. In the equilateral triangle ACB, let fall the
perpendicular CD, which bisects both the angle A ACB and the base AB.
14 TMG ON OILETHY.
New, ~T7t. — sin -4CD = sin 30°, but AD — 2
sin 30° = 5^ 1
AG 25
cos 30° = 7(1 - sin2 30°) -= (1-^)== ^ but
\/3 cos 30o
v= sin 60°; sin 60° = —y-
cos 60° = sin 30°; cos 60° = ^ J
sin 30ou= 2'
cos 30° = Jj
JL „ sin 30° 2 1
tcin 30°— qf\o /o /Q' cos 30° y 3 v o 2
cot 30° = -7
^3 cos 30° 2
sec 30° =
sin 30°* 1 2 1
: \/3,
cosec 30°
cos 30° V3_ VS' 2
1 1 = sin 30° Jf
2
sin 60° = ^^7
cos 60° = -,
V3 . sin 60° 2 tan 60 = ^7\o==—i—^ \/ 3,
cos 60° 1 v 1
"2
Cot60° = C-^ = A = -l; sin60o /v/3 .s/3
"2
sec 60° = —2, cos 60° 1 ' ~2
1 1 2 cosec 60°= sin 60° _v/3 -/S'
"2
"When the angle is 45°, sin 45° = cos (90° — 45°) = cos 45°.
Now, since cos2 45° + sin2 45° = 1, and cos 45°= sin 45°,
TRIGONOMETItY. 15
we have 2 cos2 45° = 1
cos2 45° = ^
or, cos 45° = >"
nd therefore, sin 45° = >
and since sin 45° = cos 45°, sin 45°
tan 450 = 33715"°=i' , Aen cos 45°
cot ==*snr45°==
sec 45o = c^V° = x= y2' ■v/2
cosec 45° = s-5^3 =4^=72.
n/2
The following results the student should commit to memory: cos2 A + sin2 A = 1,
cos A = v(l — sin2 A); sin A = V(1 — cos2 A), sin A 1
tan A = 330 ~^tA= ^sec'A - ^' cos A 1 0
cot A =HO - ta^z = ^(cosec A -1)'
sec A == co72 = ^C1 +tai12
cosec ^ = ^-2 = </(l + cot2 ^)»
vers A = 1 — cos A.
Examples. 4
Ex. 1. If tan A = g, compute the sine and versed sine.
16 TKIGONOMETEY.
fiill A 4 cos A 3
3 sin A = 4 >/(! — sin2 A), 9 sin2 A = 16 — 16 sin2 A,
25 sin2 A =16, 5 sin A = 4,
sin A = -=1
o
cos^ = V(l-sin2^)=,^/(l-||) - ~ B'
3 2 vers A = 1 — cos A = 1 — r — r- o 5
Ex. 2. 6 (sin ^)2 = 5 cos Ay find sine, cosine, tangent. 6 — 6 cos2 A = 5 cos A, 6 cos2 A + 5 cos A = 6,
5 cos2 A 4- - cos A = I, 6
2 ^ , 5 ^ , 25 1 , 25 169 COS2 A + 2 COS A + 777 = 1 + =-T-r = TT-:' ' 6 1 144 1 144 144
># . 5 13 cos A + 12 = ± 12'
># 2 3 COS A = g or — 2'
sin ^4*= ^/(l — cos2 (l — § = "g"'
ys . sin A 3 sj 5
tan A=^l~ T -1"'
/g Ex. 3. sin A cos A = -j-. Eind sin, cos, and arc.
• o vx V'3
sin ^4 a/(1 — sin2 ^l) =
TRIG ONOMETE, Y. 17
O sin2 A (1 — sin2 A).r=jff
sin4 A — sin2 A = —
sin4 A — sin2 A + ^
.o.l 1 sm- A ^ = — 4'
• , , 3 1
sin- -4 = 4. or | >
• , \/3 1 ' s1n^ = ^->or-.
= 60°, or 30°.
cos A = 01
4 Ex. 4. tan A + cot A = —r^, find tan A, and cot A.
v "
tan A tan A s/&
4 tan2 A — —777 tan A — — 1,
v o
4. •> ^ 4 , .41
2 1
tan A = \/3j or -77: > V d
cot or
Ex. 5. If 3 sin + 5 ^3 . cos yl = 9^ find A. 5 -v/3 . -/(I — sin2 -4) = 3 (3 — sin ./I),
25 . 3 . (1 — sin2 -4) = 9 (9 — 6 sin A + sin2 A),
18 TllIGOXOMETEY.
25 = 25 sin2 A = 27 —18 sin A + 3 sin2 A, 28 sin2 A — 18 sin A = — 2,
. . . 9 . . 1 sin- A — Y4 sin A = —
. , „ 9 • , 81 _ 81 —56 _ 25, sm A — -L4 sin A + 784 784 784
. , 9 _±5 sm A 28 '
. , ±5 + 9 1 1. sm A 2' 01 7 '
-4 = 30°. 1 + n/3
Ex. G. sin A + sin i? =
. . . „ ys sm Asm Jf = ■ Find and ;
. _ 1 + 2 v/3 + 3 sin2 A + 2 sin A sin B + sm21? = -g >
. > . ^ 4^/3 4 smr A sm Jj =
sin2 A — 2 sin A sin JB + sin2 jB =
sin A — sin £ =
sin A + sin £ =
T' 1 — 2^3 + 3
4 V3 - 1
2 ' 1 + '\/3
2
2 sin = 1; sin A — j-; A = 30°,
2 sin i? --=■• V3 ; sinJ?=^^; .•.^ = 60°.
Ex. 7. Shew that sec2 A cosec2 A — sec2 + cosec2 A. sec2^ cosec2A = sec2 ^ (1 + cot2 A) — sec2 A + sec2 ^ cot2
1 cos2 A 1 =scc"^ + wFa "ssra=scc' ^ + siso:
= see2 A + cosec2 A.
TEIGONOMETEY.
Ex. 8. If tan2 A+4: sin2 A = 6, prove that A = 60° sin2 A . . . o j r sin2 ^ , . . 2 . . —5—5+4 sin- ^=6, or, ^^+4 sm2 A=6 cos2 A 1 — sm2.4
sin2 A+4 sin2 A — 4 sin4 A = 6 — 6 sin2
4 sin4 ^ — 11 sin2 ^ = — 6, 11 . 2 . , 121 121 96 25
sm 4 ^ sin2 A 4 ' 64 64 64 64'
• 2 j 11 5 . 2 , 11 , 5 n 3 sm2 — —— h— : sin2^4 = —-4--r=2, or- ; 8—8 8—8 4
/g smAz=z— =:sin 60°; .'.A=z60O'
The *72 being greater than unity is inadmissible.
Ex. 9. * If sin x cos 03+0 sin2 cczrJ, find z. sin# \/(l—sin2 cc)=3—a sin2 x, sin2 x — sin4 x — b2 — 2ab sin2 % + a2 sin4 z, (a2 + 1) sin4 x — (2ab + 1) sin2 x= — 52,
sin4 x - /2ab+l\ .
— I ——-—- 1 sin4 .r — — \a' 2+1jsmx— a2+1'
AaW-^Aab-^-l P 4(o2+l)3 <?+!
4a5+l —4i3
4 ((i2+l)2 ' 2fl5+l n/(4^+1—4^)
2 (a2+l) 2 (o3+l)
sill2 ^ = 2fli+l±y(l+4a} —4P) 2 (a3—1) '
o;n f2g»+l±y(l+4fl8— 433)1 » t ^™.2(«3+l) J •
Ex. 10. 25 sin.il (sin^4—cos A)=4.. Prove that sin^l
4 sin? A — sin A cos A = — >
Jib
20 TniGONOSIETliY.
4 sin2 A — — = sin A cos A
25 = sin A \/(l — sin2 A),
• I J 8 • o , t 16 . , . ... sm4 A — sm- A -4 = sm2 A — sm4 A, 25 625
i ^ 33 • o , 16 2 sm4 A siir A— > 25 625
. w 33 . „ . 8 sm4 A sirr A= , 50 625
... 33 / 33 u 1089 8 961 sm 50 81112 A + Vioo/ —looob-625—10000'
. , , 33 31 SUL" d. = ■+■ 100 — 100'
. „ , 33 31 64 , . x. ' sin Iobd::iob=roo' g upper Slgn'
• j-8-4 ..smA—l0 5-
Ex.11. 6 tanu4+12 cot A = 5^3 . sec^. Findtan^. 1 o
6 tan A+, -=5^3. y/(l +tan2 A), tan A
144 36 tan2^ + 144+—--=75 + 75 tan2^, tan2 A
39 tan4 A — 69 tan2 A = 144, 13 tan4 A — 23 tan2 A =z 48,
, 4 . 23 „ 48 tan A — -— tan- A — -—>
13 • 13
tan4^-r3tan2^+©2=G- 529 2496 3025 C76 ' 676 676
, „ - 23 55 tan- A — —in-+- —j 26 —26
A , , 23 55 78 tan2.4= H— = —=3 :
26—26 26 ' .\tanJ=\/3; ^=60° (p. 14).
TEIGONOMETRY. 21
Ex. 12. If m = tan A+sin A, and n=2tanA—sm A, find an equation involving only m and n.
m = tan A + sin A, n = tan A — sin A,
m+n . j m—n tan A = —X., sm A =
j tan A sm A = — ; ^(1 + tan2 A)
m-\-n m—n_ 2
™=7HWj'
(m — 4 (m+n)2
4 (w+?^)2~~4+(m4-7^)2,
1 + 4 4 (wi — w)2+(m — n)2 (?;i+^)2 =4 (w+w)2
(w2—w2)2=4 (7?i+?2)2 — 4 (m —w)2= 16wm, m2 —. w2=: \/( 16 wzw)=4 v/ (ww).
Ex. 13. If a (sin 0)2+3 (cos &f=m (1), ^(sin0)2+«(cos0)2= n (2),
-and a tan 0=5 tan 0 (3),
«, 1 1 1 1 then —_-j #
a b m n a (1—cos2 0)+b cos2 0 = m}
a — a cos20 +1 cos2 6 = ?«, (a — b) cos2 0=a — m,
0/, a — m o a — b n a — b cos20= .-.800-0== orl+tan20= ;
a — o a — m a — m , a — b m — b , .-. tan2^= 1= .
a — m a — m In the same manner,
4. n—a tan- 0=- ; o — n
'22 TBIGKHfOltETEY.
. taii2'^ m—bb-^n tan20 a— m' n—a
but by equation 3, tan Q b tan2 0 b2
tan0 a ' tan20 d2' 52 rn—b b—n bm—b2—mn-\-bn
* * a2 ^—w • n—a an—mn—a2-\-am abhi—b^mn—cPW+ab2in=a2bm—arb2—armn+cPbn„y
. *. (d2—b2) mn=a2bm—ob2m -^aHn—ab^n — ab (am—bm+an—bn) =ab {(a—b)ni*{-(a—b)n};
(«+^) mn=ab (m+n), . a+b
ab mn 9
1111 1— a b m n
(1) Prove that . tan A cos A tan A cot A
sm A=.- —— = —, sec A cot A cosecA j sin A cot A sin A oosec A cos A =z = -= ,
tan A cosec A sec A
tan A— cos _cos A sec ^_sin A cosec A ~~sin A cot2 A cot A ~~ cot A V
coj. ^ cos A sec A sin cosec _ sin A tan A tan A cos A tan2 A9
j tan A sin A cosec A tan A cot A SCO -O, —; , sm A cos A cos A 9
A j sec A cos A sec A tan A cot A cosec ^1=—.—7 — — tan A sm A sin A
(2) cos2 A cot2 A= cot2 A—cos2 A. . . cos2 A — cos2 .Z? (3) t—j —-=tan-JB-taii2^. v ' cos2 A cos2 M (4) If»i=cosec^—sin A; and n=sec A — cos A;
then ma n%-\-m%n* = \.
TUIG ON OMETRY. 23
(5) Given l find A. A=45Q. cos A
(6)» sin A cos ^=~^find A. A=:30o.
' (7) tan ^f+cot ^=2, find A. ^1=45°.
(8) "Write down the sum of all the exterior and also of all the interior angles of a polygon of n sides ; and thence deduce the value of each of the vertical angles of the trian¬ gles made by producing both ways all the sides of a regular polygon, and verify the result in the cases where n—3 and w=4.
(9) 7OT+ a/(1 -sio)=sec A'find sec A ■■
A 1 + ^5 sec A-=z—L_r— 2
(10) s/(tan4 A—l)+ ^ 1 —t'^-2)+1=sec2 A> find
tan A;
(11) sin3^l+icos2^=l, find A. 2i
4=90°.
(12) sin4 —2 sin2 A—1=2 sin —cos2 A, find^/. A=270o.
(13) tan4 A—tan2 -4+1=4 tan A+seo? A) find tan A. (14) If sin A-\-cos Il=a,
sin ^+cos A—l,
- 1 )•
(15) Given tan x——-—1——i— +3, find tan x : v y tanas ■y/tan® 7+^13
tan a = .
24 TEIGONOMETET.
a3y\ tfy\-
tan x—tan ?/=2 ^ tan x=3,
—(tan cc4-tan v)= V 12 tan x tan ?/J tan z/=zl.
(16) Given tan x tan y=tan2 x — tan2«
2tan^=tanV+tan^ i tan x—tan^ /
Find tan x and tan y, without quadratics,
tan^V^L5; tan V=^-.
(17) tan x—tan y=2 ^ tan £c=3,
"J
(IB) 1 tan2 x = -— ; find tan x: 4 tan x — 1
tan x=z2t
(19) tan2 ce+48 cot2 cc=4 cot #+18,
tan x=4: and—2. 4 sec2 x
(20) —-—=65 cot x—39 ; find tan x: 3
tan x—-. 2
3 (21) sin2cc+sin J find x and 3/:
f
sin x sin ) ccz=:90o and y=:30o.
(22) —tan3x~S~ — tana:; find tanx:
tan x=2.
(23) If tan"2 A—cot A—- (cot A-^~ 1), ^=45°. 8
(24) (tmA — I)2—8 cot^4(1—2cotA)=zS: find tan A«
(25) Prove that
-1 =r-{ y/(scc2^ + 3) — tan A\. tan A +tan ji+&c.,ad inf.
TRIGOXOMETEY. 25
CHAPTER II.
7. To find the sine and cosine of the sum and difference of two angles. B
Let the angle FA C=:A, and the angle / FAP=B, then PA C=A +_Z?, draw 1>F p / perpendicular to AF\ and irom P and F /j\ let fall the perpendiculars PD and FF on / 0l~^iF
A G} and draw FG parallel to A C. Now, / ^\ because GF is parallel to A C, the alter- | ) nate angles, AFG and FA C are equal, a — d' e G
but FAC=A, therefore AFG=A; and since AFP is a right angle, the angle PFG is the complement of AFG; but FGP is a right angle, therefore FPG is the complement of PFG, that is, the angles FPG and AFG are the complements of the same angle (PFG), they are therefore equal; but AFG has been proved equal to A, hence FPG—A.
. , . PG+GP FF+PG
_FF PG ~APJrAP __FF AF PG PF ~AF' AP+PF' AP5
FF bat by the definitions, -j-p=sm A,
, AF PG and^-^CQS i?; also ^^icos i^Ptf=cos A,
PF and—>= sin P, hence
-/LJr sin (u4+i?)=sin A cos 5+cos A sin B. . AD AE—DE AE GF
C" ^+i,)=2p=^p-=S.-22.' (because GFzzzDE)
_AE AF GF PF ~~AF ' AP^PF ' AP =cos A cos P—sin A sin P.
c
26 TEIGONOilETRY.
To find the sine and cosine of A—JJ. Let the angle DAF—A, and the angle f /
PAFz=:2}, then L PADz=zA—JJ: from any /\\ point JP draw the perpendiculars PF and P JPD on AF and AD, also draw JPG parallel I \ to AD, then Z- Pi^^^the complement of i-J AFE, and therefore=u4, because AFG is A
the complement of A ; and because GD is a rectangle, ED=PG, and PD= GE.
. , , ^ PD FE—FG FE FG S'ii }~AP~ AP ~AP AP
_FE AF_FG IT ~AF'AP FP'AP =sin A cos B—cos A sin B,
, J ^ AD AE+EB AE,PG cos (A—Bj-Jp- —Jp —Jp^AP
AE -AF.-PG PF ~AF'AP+PF'AP =cos A cos i?+.sin A sin D.
8. Now, since sin (A-{-D)=sm A cos cos A sin B, if we make D=A, then sin {A+A )=sin 2 A=sin A cos A+cos A sin ^A=.2 sin A cos A; also, since cos (^+i?)=cos A cos D—sin A sin D; make B—A\ then cos (A-\-A)=cos 2 A=cos A cos A—sm A sin A
—cos'2 A— sin2 A; but since cos2 A = 1—sin2 A, we have also
cos 2 A—l—sin2 A—sin2 A=l —2 sin2 A ; also sin2 A—l — cos2 A;
cos 2 A—cos2 A—(1—cos2 A)=2 cos2 A — 1. Since A may be any angle, we have the sine of any angle
equal to twice the sine of half that angle multiplied by the cosine of half that angle; and the cosine of any angle is equal to the square of the cosine of half that angle minus the sine square of half that angle. For the sines we have
TEIGONOMETEy.
• A
sm A=2 sin — cos
• 1 , . • 1 ^ 1 / sin -A=2 sin - A cos- A,
. 1 , o • 1 , 1 / sm - A =2 sm - A cos - A,
sin -=2 sin A cos A. n 2 n 2 n
In the same manner for the cosines, cos 2 A=\—2 sin2 A,
cos A= 1—2 sin2 ^4,
cos ^A=-l—2 sin2 ■-A,
cos ~ A= 1—2 sin2
3 6 '
cos - A=l—2 sin2 — A. n 2 n
Also cos 2 A=2 cos2 —1,
cos A=2 cos2 —1. 2 '
cos ^ A—2 cos2 -^1—1, 2 4
cos ^ ^1=2 cos2 —1,
cos - A~2 cos2 —A—1. n 2 n
9. To find sin 3 A. sin 3 A= sin (2 A-\-A).
By putting 2 A for A, and A for £ in the sin (A-^JB), we have
sm (2 A-\-A)=sm 2 A cos A + cos 2 A sin A ; c 2
28 TMGONOMETHY.
but sin 2 A = 2 sin A cos A, and cos 2A = l — 2 sin2 A sin (2 A + A) = 2 sin A cos A cos A + (1—2 sin2 A) sin -i,
= 2 sin A cos2 A sin A — 2 sin3 ^4, = 2 sin A (1 — sin2 A) + sin A — 2 sin3 A}
= 2 sin A — 2 sin3 A + sin A — 2 sin3 A, = 3 sin A — 4 sin3 A;
that is, sin 3 A = 3 sin A — 4 sin3 A. To find cos 3 A. cos 3 A = cos (2 A + A) — cos 2 A cos A — sin 2 A sin
= (2 cos2 A — 1) cos A — 2 sin A cos A sin A, = 2 cos3 A — cos A — 2 cos A sin2 A, = 2 cos3 A — cos A — 2 cos ^ (1 — cos2A))
— 2 cos3 A — cos A — 2 cos A + 2 cos3
= 4 cos3 A — 3 cos A. In the samp way tvg may find the sin 4 A, &c., but it
will be better to express them in general terms, as in the next Article.
10. Now, since sin {A + JB) = sin A cos H + cos ^ sin -#j sin (.A—i?) = sin A cos JB — cos A sin U,
by addition we have sin (A + J3) + sin (A — i?) = 2 sin A cos B (1).
By subtraction, sin (A + B) — sin (A — JB) — 2 cos A sin £ (2).
By multiplication, sin (A + jB) sin (A — 7?) = sin2 A cos2 £ — cos2 A. sin2 i?
= sin2 A (I — sin2 JB) — sin2 JB (1 — sin2 A) = sin2 A — sin2 JB = (sin A + sin i?) (sin A — sin _Z?).. .(3).
Also, cos (A + JB) = cos A cosJB — sin A sin JB, cos (A — JB) = cos A cos JB + sin A sin JB.
By addition, cos (A + i?) + cos (A — D) = 2 cos A cos JB (4).
By subtraction, cos (A — I?) — cos (A + JB) = 2 sin A sin JB (5).
TKIGONOMETEY. 29
By multiplicalion^ cos (A + B) cos (A — i?) = cos2 A cos2 It — sin2 A sin2 B
= cos2 A (1 — sin21?) — sin2 £ (1 — cos2 A) =z cos2 A — sin2 £ = (cos A + sin 2?) (cos A — sin 1?) (C).
Now by equations (1) and (4): sin (A + JB) + sin (A — i?) = 2 sin A cos B, cos (A + J?) + cos (A — JB) =z 2 cos A cos JB ;
br B put dj and for A put na; then, A-\-B =na + a = (n + 1) a, and A—B = na—a={ji—1) a,
sin (m + 1) a + sin (w — 1)0 = 2 sin na cos cos (n + 1) a + cos (« — 1 a = 2 cos wa cos a.
Ey giving to n different values, the sine or cosine of any nultiple arc may be found.
11. EW, since half the sum of any two quantities idded to half their difference gives the greater, utA half ;he difference subtracted from half the sum gives tl e less,, ve have
A=i(A + B) + i(4-l?), B=:i(A+B) - i(A-B).
Putting -b(A + B) for A, and %(A — B) for By in sin A + B) and cos (A + B);
sin A = sin {i(A + B] + ^(A — 7>'} =sin%(A+-5)cos%(A — B)+cos%{A-^-B) sin %(A — B)..(l).
In the same manner, (in.Z?=sin£(^+i?) cos%(A—B)—cosJ(^+^)sini(^—B)(2). By adding (1) and (2), sin^ + sini? = 2 sin J( + ) cos £(A—B) (3). By subtracting (2) from (1), dnA — sini?=:2 sin-J-(^l — B) cosi(^+^) (4). jos = cos{£(A + B) + i-(A — B)} =cosi(A~i~B) cos%(A—B)—sin %(A-\~B) sin i(A—i?)..(5). 20s B = cos{i(A+B) — i(A-B)} =icosb(A+J3) cos i(^-11) + sin i(A+B) sini(A-B)..(6).
30 TEIGOTTOMETEY.
Add (5) and (6): cos A + cos B = 2 cos %(A + B) cos —B) (7). Subtract (5) from (6) : cos B — cosA = 2 sin ±(A + B) sin i(A — B) (8). By dividing (3) by (7) : sin A + sini? __ 2 sin %(A + B) cog i(A—B) __ sin %(A + B) cos A + cos B u 2 cos i(A + B) cos £(A—B) ~ cos £(A + B)
= tan %(A B) (9). Divide (4) by (7): sinA—sinB 2smi(A—B)co8i(A+B) L
cosA+cosB~?cosi(A+B)oo^'A—B) -taiia^— Divide (3) by (4) : sin A + sin B 2 sin i( A + B) cos %(A — B) sin. A — sin B~2 sin i{A — B) cos |(A + B)
tan i(^+-5) — ^ l;'
Divide (7) by (8): cos B + cos A_2 cos %(A + B) cos £(A —B)
-cos B — cos A~~2 sin £(A + B) sin A — B) — coti(A+B) coti(A—B) (12).
JMvide (4) by (8): sin A — sin B 2 sin $(A — B) cos £(A + B) cos B — cos A 2 sin i(A -f B) sin %(A — B)
~ cot i(A 4- B) (13).
12. To express the tan (A ± B) in terms of tan A and tan B.
sin (A + B) _ sin A cos B + cos A sin B tan ( + ) — COg^ +A) cos A cos B—sin A sin B'
Divide the numerator and denominator of this fraction by cos A cos B y then
sin A cos B cos A sin B , j t cosAcosB cos A cos B__ tan A + tan B tan (A + Jj) ^ sin A sin £ ~l — tanAta.xi3'
cos A cos B
TlilGOtfOMETKY. 31
f r j v\ — s^n sin AzosB — gos ^ siu ^ an ^ ' cos (A — i>) cos A cos + si11 A sin It sin A cos i? cos sin JB
cos A cos JS cos A cos M tan A—tctfii? sin A sin JJ . 14- tan^i tan ^ COS COS i/
13. To express^ the cot (A ± JB) in terms of cot A and cot B.
cos (.A + JB) cos AcosB — sin A sin J2 co • ( + ) g.n ^ ^ g-n ^ cos £ cos ^ g.^
Dividing numerator and denominator by sin A sin JB, cos.^ cos JB j
a i 7>\ sin ^ sin B cot A cot JB—1 co ( + ) ~~ sin A cos jB cos A sin jB cotM + cot A '
sin A sin JB sin A sin JB making JB =: Aj ^ve have
^ ^ cot2 A — 1 cot 2-4= 2cotA ■
In precisely the same manner we obtain ^ cot. ^ cot i? +1
cot (A —B)= cotB_cotA.
If in tan (A -p If) \re make £ = A, ,, , , a > a\ tan A + tan A tta tan (A + A) =
that is, tan 2A = ^ ^an
If for 2^4 we put A, and therefore for A, "vrc have
1 — tan2 A' A 2
2 tan-~ tan ^ J
1 — tan2 —r
In the same manner
32 TKIGONOMETRY.
tan—= 3
A 2tan-6
73' 1 — tan2 — 6
. A 2tal1^ tan — = , n 1 Ar 1
14. Inverse trigonometrical functions. The quantities sin-1 A, cos-1 A, &c., are called inverse
functions, and have the following signification :— sin-1 A is put for an angle whose sine is A, cos-1 A is put for an angle whose cosine is A, &c., &c. Let sin a = a, and sin 13 — b,
then a = sin-1 a, and /3 = sin-1 b, cos a = \/(l — a2), and cos ft = y/(l — b2),
sin (a -|- /3) = sin a cos /3 + cos a sin /3 = a y/(l — b2) + b\/(l— a2);
a -f- p = sin-1}^ \/(l — £2) + ^ \/(l — «2))> but a = sin-1 and ^3 = sin-1 b ;
sin-1 a + sin-1 b = sin-1}^ \/(l — Z"2) + ^ \/(l —■ «2) j. Also in the same manner,
sin-1^ — sin-1 b = sin-1 [a \/(l — b2) —b \/(l - a3)}. By proceeding in the same manner for the cosine, cos-1 a + cos-1 b = cos-1 [ab — \/( 1 — a2) — b2)}, cos-1 a — cos-1 b = cos-1 [ab + V(1 — a2) y/{\ — ^2)j.
Proceeding in the same way for the tangents, if tan a = fl, tan ft = b,
then a — tan-1 and ft = tan-13, , , , ^ _ tan a + tan ft _ a + b tan (« + p) — j _ t,m a tan ^ l—ab>
(a + l\ •••» + /»-tan-
but a = tan"1 and j3 = tan-1 by
TIUGONOaiETKY. Kj0
f (i -f- b \ tan"1 a + tan-11 = tan ^ _ ai)<
s tan a — tan ft a — b and since tan (a — ft) = i + tan a tan ft ~ 1 + ab'
/a — b \ tan-1 a — tan-1 b = tan-1 ^ ^ J-
In the same way,
cot-1 a -+- cot-1 b = cot-1 (a^ 1Y — \b ± a )
15. The following formulae, which are sometimes called the formulce of verification, are very useful l—
since cos2 A + sin2 = 1, and 2 sin A cos A = sin 2A.
By adding and subtracting we have cos2 A + 2 sin A cos A + sin2 A = 1 + sin 2Aj
and cos2 A — 2 sin A cos A + sin2 A = 1 — sin 2A. By extracting the square roc t of each of these,
cos A-\-smA=: \/(l + sin 2A)j cos A — sin A = \/(l — sin 2A).
When A is less than 45°, sin A is less than cos A. By adding,
2 cos A z=z >/(! + sin 2A) + \/(l — sin 2A),
"cos A = ^ \/(l + sin 2-4) + ^ \/(l — sin 2-4).
Subtracting, 2 sin ^ = \/(l + si11 2.4) — ^{1 — sin 2-4),
sin A = - VC* + sin 2^) — - — si11 2-4). Jh 2
Wh£n A is greater than 45°, then sin A is greater than cos A, and the above will become
cos A =- ^ {(\/l + sin 2A) — V(1 — sin 2^)},
sin ^ = \ {\/(l + sin 2-4) + — sin 2A)}.
To find tbo sin 18°. c 5
34 TUIGOXOMETnr.
Since the cosine of any angle is equal to tlie sine of its complement,
cos 54° = sin 36°, but cos 54° = cos (36° +18°) = cos 36° cos 18° — sin 36° sin 18°;
cos 36° cos 18° — sin 36° sin 18° = sin 36°, but cos 36°= 1 — 2 sin2 18°, and sin 36° = 2 sin 18° cos 18° ;
cosl80(l — 2 sin218°) — 2sin218°cos 18° = 2 sin 180cosl8c'. Dividing by cos 18°,
1 — 2 sin2 18° — 2 sin2 18° = 2 sin 18°; .\ 4 sin* 18° + 2 sin2 18° = 1,
sin2 18° + ^ sin 18° = ^
1 . /1\2 11 5 sm2 18° + 2 sm 18° + =g+ 16 = 16'
1 V5 sm 18° + 4 = -4 ,
sin 18° — —5 - . 4 The same result may be obtained by using the formula
for thexsin SA; or by the 10th proposition of the 4th book of Euclid* cos 18° = - sin218°)
_ /5+V5 V(5+V5), ""V 8 2\/ 2
sin 36° = 2 sin 18° cos 18° _ 2(^5-1) V(5 + V5) (y/S - 1) V(5 + V5)
4 ' 2V2 — 4^/2 ^{(5 - 2^5 + 1) (5 + ^5)1 _\/(20 - 4^5) ~ 4-^2 4
1 /20 — 4 y/b _ 1 = 272- V 4 272-V/(5-\/5).
Inow the sin 36° = cos 54°, •
TRIGONOMETRY. 35
cos 54° = VC5 — n/ 5)>
sin 54° - ^(1 — cos3 64°) = /y/(l — 5 ~8 ^
/8 —5 + ^5 /3+y5_ /6 + 2V5 — V 8 = V 8 "V 16
>^(5 + 2 ^ 5 -fr-1) \/5 + 1 = 4 "" 4 '
Or thus, sin 54° = cos 36° = cos 2 (18°) = cos2 18° — sin218° 5 + ^/5 6 — 2^/5 10+2^5 — 6+2^/5
8 ~~ IG- "" " 16 4+4^/5 1+ ^5
16 — 4 ;NW since sin 18° = cos 72°, and cos 18° = sin 72°, we
have the following results:
clS'-^+Z5',
™361" ; "■36- - S
■ N/S + 1 */(5 — */5) slno4o=___. cos 54° ~ g a/2 '
sin 72° = y(52y
5) ; cos 72= = .
By the formula, page 34, we may find the sin 15°.
sin 15° ^{^/(i + gin 30°)_ ^(1 _ sin 30°))
=H a/O+§) - { A/O " §) ) 1 /V3 1 \ V3 —1
~2W2 v/2/- 2^2 • This might have been found by putting
sin 15° = sin (45° —30°)
36 TllIGOXOMETIir.
Examples. (1) Shew that t
cos (60° + A) + co§ (60° — A) = cos A. cos (60° + A) = cos 60° cos A —* sin 60° sin A, cos (60° — A) = cos 60° cos A + sin 60° sin A.
By addition,
cos (60° + ^) 4-cos(60o — A)= 2 cos GO0 cos.4 = 2 X % <&sA
cos A ^because cos 60° =
(2) sin (30° + A) + sin (30° — A) = cos A. sin (30° + A) = sin 30° cos A + cos 30° sin A,
* sin (30° — A) = sin 30° cos A — cos 30° sin A. T3y addition,
sin (30° + ^) + sin (30°—A) = 2 sin 30° cos A = cos A. 2 tan -I A
(3) sin A = i
2 sin cos sin -4 2 sin -^4 cos ^ cos2 iA
sin A = ' 1 "" 1 """ 1 cos* ^A.
(by dividing numerator and denominator by cos3 bA) 2 tan %A 2 tan -J- A
= sec2 iA " 1 + tan2 ' (4) 2 cos (45° + ^A) cos (45° — %A) = cos A.
cos (45° + ^4) = cos 45° cos &A — sin 45° sin j?A 1
= *^2 (C03 ^ 5
cos (45° — iA)= cos 45° cos ZA + sin 45° sin ^A 1
= ^7 (cos ^A + sin i-A).
Ey multiplication, 2 cos (45° + iA) cos (45° — iA)
= 2 X ^ X 72 ^cos2 ^ — s^112
= cos2 —sin2 -hA = cos A.
TEIGONOMETBY. 37
(5) Given sin {x -{- a) + cos (x + a) = sin (x — a) + cos (.r — a), find x.
sin (cc -f a) — sin (cc — a) =: cos (cc - a) — cos (x + a), sin {x -f a) = sin x cos a + sin a cos x}
sin (x — a) = sin x cos a — sin a cos x ; sin (x + a) — sin (x — a) = 2 sin a cos x, cos (a; — a) = cos x cos a + sin x sin a, cos (x + a) = cos x cos a — sin a; sin a ; cos (x — a) — cos (x + a) = 2 sin x sin a; lience, 2 sin a cos x = 2 sin a sin x;
cos x = sin x; x = 45°.
(6) If sin x + cos 2 x = 5 Prove a? = 18°.
Since cos 2 # = 1 — 2 sin2 sc, we have
sin ^ 1 — 2 sin2 x = ^5,
1 . 2 — -v/5 sinw x — - sin x-— — ; 2 4 '
1 . 1 2 — . 1 8 — 4^/5 1 sm** cc — - sin a; + ^ = = 2 16 4 16 16 ^ 16
_ 9 — 4 s/5 _ (\/5 — 2)2
16 16 ' 1 s/5 - 2 sin cc — - = — , 4 4 ' \/o — 2 . 1 \/o—1 . -oC sm a; =z + - = — = sm 18° ; 4 4 4
x = 18°.
(7) Find the tangent of 15°.
4. ico <- fAzo qaon tan 45° - tan 30° tan 15° = tan (45°- 30°) = •—— j^r ^3 v 7 1 + tan 45° tan 30° 1
1 ys ^ s/3 - 1 = (N/3 — I)2
. , 1 \/3 + 1 2 + ^3
2 — v/3.
38 TRIG OX03LET11T.
(B) If x = sin-1 and y = sin 1 then, x + y = 90°. 5 5 3 / r /syi 4
Smee sm x — ^ ; cos x — a / i l — I g ) r = g,
andsiny coS,j = ^/ {l — 0) | = |,
sin (x + y) = sin x cos y + cos x sin y _ 3 3 4 4 __9_ i6_25_
5^ 5 5 X5 25 ~^2o 25 ' but sin 90° = 1; x y = 90°,
or we may put it in the following form,
. , S . , 4 . 1 /3 3 ,4 4\ . , , S1 5 + Sm 5 = Sm \5><5 5X5/ = 8111 1- (9) Shew that 4 sin 9° = ^(3 + s/b) — a/(5 — */b).
By Art. 21, sin^ = g{x/(l+ sin 2 A) — */(! — sin 2 A)]j
we use the under sign as the angle is less than 45°. Now, in the above formula let A = 9°, then, 2^ = 18°,
sin 9° = ^ {\/(l + si11 IS0) — VX* 180)}
= 5{V(3+y5)-y(5-y5)};
4sin90 = y(3 + v'5)—y(5—/S). (10) Shew that 2 cos 11° 15' = ^{2 + ^(2+^2)}.
Since 2 cos2 ^ d. — 1 = cos A, /I + cos A
cos f A. = a / 2 '
let A = 45°, then J A = 22° 30',
/I + cos 45° A/1 + 72 cos 22° 30 = a / 2 — 2
7^2 + 1 /2 + V2 — V 2^2 ""V 4
n/(2+>/2) 2 '
TEIGONOMETHy . 39
< ^ 1 _ n/(2 + v/2)
_ , , /I + cos 22° 30' ' cos 11° 16 __ 2 —- 2
/2 + \/(2 + \/2) Ay{24-\/(2+ \/2)} #
- V 4 ~ 2 2 cos 11° 15'= v/{2 + 7(2 + ^/2)}.
Or it might have "been put in this form: o450 , _ 1
2 cos- 1 = cos 4o0 =
45° 2 2 (2 cps2 1) = ^2 — ^
45° 4 cos2 — = 2 + \/2,
45° 2 cos — = ^(2 + ^2),
45° 2(2 cos2 — -l) = >/(2 + >/2);
j 5° 2 cos — = V{2 + 7(2 + ^2)}.
45° This is the same as 2 cos — — ^{2 + ^/(2 + n/2)}, and
45° by continuing the operation to —, we have 2n
45° 2 cos — = V {2 + ^/(2 + to w + 1 terms)}.
(11) If A, JB, and C be in arithmetical progression, sin A — sin B = 2 sin (A — JB) cos H.
Since J3 is an arithmetical mean between A and C, we have i{A+C)=£ (1), i (A - C) = JB ~ C (2),
A—B =B-C (3). By page 29,
sin A — sin (T = 2 sin % (A — C) cos J (.A + C) = 2 sin (B — C) cos B; by (2) and (1) = 2 sin (A —B) cos i?; by (3)
40 TEIGONOilETRY.
(12) Show that
l! 3 1 3
tan 1 - + tan 1 ^ + tan-1 ^ + tan""1 g == 45°.
tan 1 o + tan 1 ^ + tan"1 ^ + tan-1 |
= ^an 1 ^ + tan-1 ^ + (tan-1 y +* tan""1
f 1 , M 1 1 I S ' 5 I 4
tan"1 - + tan-1 g = tan-1 i —- > =- tan-1
l1 — s'si
1 T 8 T-ir=taii_1 if; tan-1 ]j + tau-! ^ = tan-1 ' 7
= tan"1 (1) = 45°;
I'-rsJ . •. tan"1 ~ + tan"1 \ + tan"1 \ + tan"1 \ = tan"1 ~ + tan"1 —'
o o 7 o 7 11
f - + - but tan"1 ^ + tan"1 ^ = tan"1 j ——4~^"
tan"1 \ + tan"1 \ + tan"1 1 -j- tan"1 i = 45°. o o 7 o
CC 1)»Xj 'y* (13) Given vers"1 rers"^! — £) = vers"1 —; find-
ct (Id
cos-1 — 0 — cos-1 {1 — (l _ })} = cos-1 (l — j)>
T+T {''"l1 —«)}'
-(-3'='- 1 — l _ 25 T+T — T+l'
25 1 « ~ ± V 1 + b '
TRIGONOMETRY.
. x—i — / 25
(14) If sin x = -^and cosy= ^72 + 2^3' then#—y= 30°.
1,1 \/0 -{- 1 cos y = 1 1— V2 2^/3 2 \/3
co"= a/C-s)-"/? sin (a: — y) = sin x cos 1/ — cos a: sin jr
_ 1 ^6 + 1 ^2 ^(5 — 2^6) — Vs' 2 ys ^3 ' 2 V3
_v/6 + l v'2.n/(5 — 2^6) — 6 6
_ ^6 + 1 — n/CIO — 4 V6) _ yo + 1 — ^(4 — 4 yo + 6 "6 C
N/6+l-y(N/6-2)2 y6+l-(v/6-2) 3 1 . 6 = - = - = sm30°
Hence x — y = 30°. (15) If 4 sin x sin 3x = 1, find
4 sin 57 (3 sin a; — 4 sin3 #) = 1, 12 sin2 cc — 16 sin4 x — 1,
•4 3 . „ 1 sin. x — ~ snr x = — —, 4 16 .4 3 . 0 9 9 1 5 sm* cc — - sm~ x A — —,
4 64 64 16 64 • .> 3 a/5 snr cc — - = -+- 1—
8 — 8 3 dz \/5 6=1=2^5 (75 ± I)2
Sm X 8 — 16 — " 16 ' n/5 ± 1 sin x taking the upper sign, x = 54°,
and taking the under sign, x = 18°.
42 TEIGONOMETRY.
(16) If cos x = and cos y = > show that
x + y = 60°.
V(v/6— I)3 V6 — 1 — 2 -v/S — 2 Ts"'
cos (a; -f- if) — cos x cos y — sin x sin y
_ /2 ^3+^/2 1 VO —1 ~ V 3 * 2^/3 — V3 ' 2 v/3
j/6_+2 /6-l_-t/6+2-V6 + l _ 3_1_ ' 6 6 — 6 ~ 6 2— '
(17) If tan 2x — -—^ tan a; ^ s]10w t}lat; the radius of the ^ 1 — 4 tan- cc
arc 57 is - • 2 2 tan a;
Since tan 2x = 1 — tan2 x 2r2 tan x , _.
tan 2a? =: -2 1—to radius r; r2 — tan- x 7
2r2 tan x 2 tan x r2 — tan2 x 1 — 4 tan2 x r~ 1
r2 — tan2 x 1 — 4 tan2 x r2 — 4^'2 tan2 ^ = r* — tan2 x
4r = 1
0 1 1 » =i; ••• ?- = 2-
(18) If sin (a + 0) sin 0 = cos2 show that 0 = 90° —
. „ a 1 + cos a Since cos- - — ,
2 2 ' / • ^ ^ 1 + cos a (sm a cos 0 -}- cos a sm 0) sm 0 =z :
2 . ^ n . 'in 1 "+■ C0S « sm a sm 0 cos 0 + cos a sm2 0 = ,
* TEIGOIsOMETRr. 43
I . . 1 — cos 20 1 + cos a - sinasin20H ^ cos a = ^ >
dn a sin 20 + (1 — cos 20) cos a = 1 + cos a, sin a sin 20 — cos a cos 20 = 1 ;
or, cos a cos 20 — sin a sin 20 = — 1; i.e., cos (a + 20) = — 1;
a + 20 = 180°;
=152^
fiO0
(19) Cos ii0 + cos (n — 2) 0 = cos 0, then 0 = - ' n — 1 '
cos {(n— 1) + 1} 0 + cos {{n — 1) — 1} 0 = cos#, cos (n — 1) 0 cos 0 — sin (n — 1) 0 sin 0 + cos (« — 1)0 cos 0
+ sin (« — 1) 0 sin 0 = cos 0; 2 cos (w — 1) 0 = 1,
cos (ra— 1) 0 — 1, 2
{71— 1)0 = 60°;
60° \ 0 n — 1
(20) Show that
sin-1 + cot""1 3 = 45°.
Let z — cot-13; cot % = 3, cos3 s 1 — sin2 s . 0 1 _ . 1
. 2 =9 or —r-5 =9; .\sin~s = — and sm s = .. sm2 z sm2 s 10
the above becomes sin-1 -i— 4- sin"1 —= 45° V10
+ ^"'"TTn = • -^7, + -L • ^g) \/5 yio Vya " vio"1" vio
dn-1 H \y5o^
(21) Show that
sin-1 (~ 1 —^ = sin-1 —— — sin""1 * 4^ Vyso^vso^ V50 V2~
tan 1 ^ + 2 tan"1 ^ = 45°. ' 3
44 _ TRIG ONOMETR Y•
This is the same as tan""1 \ + tan"1 \ + tan"1 ~ = 45°,
1 1 \7 ^/ \91 i tan*1 ^ + tan"1 - = tan"11 j—^ > = tan"1 j — = tan"1? j
then tan"1 ]■ + tan-1 \ = tan"1
Ji o
(22) Find sin A from the equation tan bA = cosec A — sin A,
sin A = cosec A — tan iA = -t-~ — tan iA. z sin A A 9
1 sin iA 1 2 sin2 iA 2sinJiLo^i cos^A 2 sin IA cos ^ A 2 sin ^A cos £A
1 — 2 sin2 jA cos A 2 sin kA cos ^A sin A 1
sin2 A = cos A =\/(l — sin2 A), sin4 -4 =■ 1 — sin2 A, sin4 A + sin2 A = 1,
sin1 A + siu2 A +1 = 1+1=1,
• , / , 1 \/5 sin- A + -= — j
. 0 j 1 . \/ 5 sm-^ =- -±-- »
^ /— 1 ± \/5 sin -4 =a/ g *
^ 2 cosec 2A — sec A (23)cot»(45° + ^)=2 cosec + sec^-
i >n (coshA — sinj--^)2
cos2 — 2 sin ^ cos ^ + sin2 %A s- cos2 ^A + 2 sin ^ cos ^A sin2 lA
THIG OK OMETIIY. 45
1 — 2 sin cos 1 — sin A
l -
1 + 2 sin %A cos £A 1 + sin A 1
cosec A cosec A — 1 cosec A + 1
s£c A cosec A — sec A sec A cosec A + sec A
2 cosec 2 A — sec A 2 cosec 2A + sec A ' 2 sin A — sin 2 A
(24> Pr0VC 2 sin A + sin czA ~ - 2 sin — sin 2 A 2 sin A — 2 sin A cos A 1 — cos A 2 sin A + sin 2 A 2 sin A 2 sin A cos A ~~ 1 + cos A
1 — (1 — 2 sin2 £A) ___ 2 sin2 %A _ sin2 %A 2 i
1 + cos2 %A — 1 ""2 cos2 l A ~~ cos2 ~~~ ^an
. sin + sin 3-^ + sin 5A (25) Prove that cos A + cos 3A + cos 5A = tan 3^.
By formula, page 29, sin A + sin 5 A = 2 sin 3A cos 2A, cos A + cos 5A =i 2 cos 3A cos 2 A,
sin A + sin 3A + sin 5A = sin 3A + 2 sin 3A cos 2^4,
~ sin 3A (1 + 2 cos 2A), cos + cos 3A + cos 5A = cos 3A + 2 cos 3A cos 2^4
= cos 3A (1 + 2 cos 2A); sin A + sin 3A -f sin 5A sin 3A (1 + 2 cos 2A) cos A + cos 3A + cos 5A cos 3A (1+2 cos 2A)
= tan 3 A, (26) If A + JB + C = 90°, prove that
cot A + cot B + cot 0 = cot A cot B cot C.
By Art. 19, cot (.A + S) = ■
46 ^ TRIGONOMETRY.
Put A B for and C for B, and we have *. t a . 7? i n\ cot (A + i?) cot 6'— 1 n cot(^ + i?+ G)- coi0+cot{A + jt) =0.
because the cot 90° = 0 ; cot (.A + B) cot C — 1=0, cot (A + B) cot C = 1, cot A cot B — 1 i n or r cot G = l, cot B + cot A
cot A cot B cot C — cot C = cot A + cot B; cot A cot B cot C — cot A + cot B + cot C.
(27) If A + B + G = 180°, prove that cos2 A + cos3_Z? + cos2 C + 2 cob A cos B cos (7=1, cos A = — cos (B + 0) — — cos B cos C + sin B sin (7, cos2A = — cos A cos B cos C + sin B sin C cos Af
cos2-Z? = — cos A cos B cos C + sin A sin C cos B, cos2 A + cos2 B ==—2 cos A cos B cos C + sin C (sin B cos A + sin A cos 5),
cos2 A + cos2 -Z? + 2 cos A cos B cos G = sin Gsin (A + B) = sin <7sin C = sin2 0 = 1 — cos2 G;
cos~ A + cos2i? + cos2 (7+2 cos A cos B cos (7= 1. On the same supposition,
sin 2 A + sin 2B + sin 2 G = 4 sin A sin B sin G, sin 2^4 + sin 2B + sin 2 G
= 2smAcosA + 2 sin B cos B + 2 sin G cos (7, but cos A = — cos (^2? + C), cos (7= — cos (A + B),
and sin i? = sin (A + G) ; .\ sin 2A + sin2B 4* sin 2 G= — 2 sin^ cos (i? + (7)
+ 2 cos B sin (A + (7)—2sin Gcos^A-^ B) —(by expanding)
— 2 sin A cos i? cos G + 2 sin A sin B sin (7, + 2 sin ^ cos B cos G + 2 sin ^ sin B sin (7, — 2 cos A cos B sin (7, + 2 cos cos B sin (7,
= 4 sin A sin B sin (7.
TIUGONOMETRY. 47
Also, sin A + sin 1? + sin 0 =" 4 cos cos -Ji? cos C: sin (7 = sin (A + i>),
sin ^ + sin _Z? = 2 sin l(A + £) cos ^(A — i?), fiin C = sin (A + .5) = 2 sin b(A + B) cos (A +.£),
i(A + 2?) = 90° - iC, ^in = sin (90° — %C) = cos |(7,
sin^+sin^ + sin 67= 2 cosj C {cosi(^ +2?) + cosi(A—£)}, but cos + B) + cos ±(A — JB) = 2 cos ^A cos -J-2?,
sin A + sin -5 + sin C = 4 cos cos Ji? cos ^(7. Or thus : since A + JB + C= 180°, then ^A + iJB + JC = 90°, sin + si11 + sin C = 2 sin ^A cos ^A + 2 sin ^B cos \B + 2 sin cos J C = 2 cos ir^f cos + J C) + 2 sin |2? sin + h O) + 2 cos J (7 cos (^A +
= (by expanding) 2 cos cos %B cos I G — 2 sin %B sin cos &A
+ 2 cos %A cos -JB cos J(7 + 2 sin sin J6' cos + 2 sin ^B sin %A cos J C — 2 sin sin ^A cos ^ C
= 4 cos cos ^B cos bC. (28) If sin 2x = (sin 3cc)2, show that a; — 15°.
Subtracting each side from unity, 1 — sin 2x = 1 — sin^Sa*, cos2 ^ — 2 sin # cos x + sin2 x = cos2 3.r,
cos cc — sin x = cos 3^ = 4 cos3 a; — 3 cos x, sin x = 4 cos a; (1 — cos2 cc) = 4 cos 03 sin2 cc ;
sin x = 0. x — 0, or tt,
and 2 sin x cos cc = 1,
or sin 2 & = i = sin 30° ;
a? = 15°.
(29) cos x + cos y = i (^2 + -/3) \
1 > , find x and y. cos 3 x + cos 3 y ———r \
48 TRIGONOMETRY.
The second equation is the same as
4 (cos5# + cos3 y) — 3 (cos x + cos y) = — -i- ; (page 28.) v 2
3 . •> M 3 j. r^3 1 \ .\ cos3 a; + cos3 y = - j -^2 H 2 >72 )'
or (cos a; + cos y) (cos2 # — cos x cos y + cos2 y)
f o 1/2^2+ 3^3\ .-. cos- x — cos x cos y + cos- y = - ^ j
2 ye + 9 — 4 - 3 ^6 , ... , . , , /0 /oN = 4 , by multiplying by (v/3 — n/2) ;
cos2 x — cos x cos y + cos2 y = - (5 — ^6),
but cos2 x + 2 cos x cos y + cos2 y = ^ (5 + 2 V 6);
3 cos x cos y — ^ . 3^6 ;
1 cos x cos y = - Q>;
cos2 x — 2 cos # cos y + cos2 y = ^ (5 — 2 sj G);
cosa: — cosy = ± ^ (v/3 — ^2) ; Ji
but cos x + cosy = - (^3 + \/2). Ji By adding these equations, we have
cos x — —, or a: = 30°, or 45°.
By subtracting them
cosy= —i or ~ y = 45°, or 30°.
TEIG ON OMETRY. 49
(1) Prove the following theorems, 2 cos A 2 tan A
sin A — cosec A sec2 v.- A tan ^ A + cot J A 1
cot £ A — cot A' sin 2 ^ cosec j- ^ — 2 sin ^ ^ cot^A — tan j- ^
cos g ^ ^ cosec -dt cot %A-\- tan J ^
1
tan ^4 =
1 + tan^tan^-^t'
2 cot % A 1 + see A cot b A — tan ^ A cot2 h A —I cot J A 7
sec2 j- ^ cosec2 j-^ __ cot + tan 2 — setf^A ~~ cosec2 ^ ^ — 2 cot ^ A — tan
cot2 % A — 1 2 — sec2 j- A cosec2 ^A — 2 C0^ ^ 2 cot £ A 2 tan %A 2 cot '
. l+tan2^-4 cot iA4-tan^A sec % A cosec li-A cosec A= —r-r—r~r = ^ =— = —-—^ — 2 tan i A 2 2
(2) cos4 A — sin4 A — sin 2 A. (3) tan (45° + A) + tdn (45° — A) = 2 sec 2
(4) tan (45° + A) — ^ tan (45° — A) = t^ji 2
cos 2 ^ _ 1 ^ ' 1 + sin 2 "" tan 2 + sec 2^' . x sin 2 ^ (6) ~^rT + cos2A=1- (7) 4 sin2 (30° + A) = 2 sin2 A + ^3 . sin 2 ^ + 1. (8) cos 2 ^ + cos 2 2? = 2 cos (^4 + i?) cos (A - B).
(9)COs^ = . secA^xeisA
(10) 8 cot 2 A cosec2 A = cot A cosec2 A — tan-^ sec2 A, (11) {cos A + y— 1 sinA— 1) (cos A — \/—l— 1)
= 4 sin2 i' A.
50 TRIGOXO^TETItY.
, . ^ sin 0 1 0 3 30 (12) Prove —r—z = - cot r — -z cot —• 1+2 cos 0 4 2 4 2 (13) Shew that tan 9° = 1 + x/5 —^/(5 +2 */5). (14) Prove that tan (45° — x) tan (45° — 3#)
1 — 2 sin 2x 1 + 2 sin 2x°
(15) Eliminate 0 from the equations cos3 0 + a cos 0 = b ; sin3 0 + a sin 0 = c.
(16) If tan 0 = tan3 and cos2 0 = g—-j o
shew that m = {(cos 0)3 + (sin 0)3
. , , NJ , « 1 — 2cos2tf n _ . (17) tan (a + .r) tan {a — x) = ! + 2 cos find sin a;,
* = 30°.
/ION TP n tan x __mtMi(a — x) , ^ ^ (18) IfcoSM«-^) shew that
n-m tana.
(19) If 4x = 3a cos 0 + a cos 30 4%} = 3^ sin 0 — a sin 30
tan (a — 2x) = —,— ^ y n + m
shew that (d2 — x2 — y2)3
= 27 drx-if. (20) If sin + x) sin (2a — /3 + x)
= (cos2 ft — cos2 a) sin (a + 2rc) — sin a (sin2 j3 — sin2 a), shew that sec cc = sin (a + ) tan (a — fi).
(21) If 2 tan-1 x = sin-12^, then y = x
l + x2
( COS 0 + COS 0 = „ , (22) Given { , ,} find cos 0 and cos 0. v J ( cos 50 + cos 50 = 0 j ^ (23) Given vers"1 (1 + x) — vers-1 (1 — x)
z=z tan-1 2 \/(l — x2), to find x,
z = ± 1,-^-and — 1.
TKIGONOaiETEY. 51
(24) tan-1 (^t)~ ^ (^fTl) =f' find ^
2 *=±,73=T
(25) sec""1 a — sec""1 b — sec11 — sec"1 -> fi11^
x = it
(26) tan""1 x — tan"1 y ~ tan""1 — tan""1 = -77, 2?/ 2cc 12
find cc and y.
(27) Prove that ^ + tan-1 (cot 2 A) = tan-1 (cot ^). (28) Find the sum of w terms of the series,
tan"1 a + tan"11 +"12a, + tan"1
1 + ° 3^ + &c.
(29) If^l + J&4- C= 180°, then (1) sin2 + sin2 Ji? + sin2 J (7+ 2 sin % A sin Ji? sin J (7— 1. (2) tan A + tan JB + tan C = tan A tan B tan C. (3) cos 2 A + cos 21? + cos 2 (7=1 — 4 sin ^ sin i? sin C.
(30) If A + B + Gz=z 90°, then (1) cos 2 A + cos 2 B + cos 2 C= 1 + 4 sin A sin B sin C, (2) tan A tan B + tan A tan C + tan B tan (7 — 1. (3) tan A + tan B + tan C = tan A tan B tan (7
+ sec A sec B sec C.
CHAPTEE III.
16. There are six parts in every plane triangle, viz., the three sides and the three angles, any three of which being given, tha others can be found, except the case where the three angles are given; for it is clear that if you draw an indefinite number of lines parallel to the base AB of the triangle ABC, all the triangles so formed will have their
d 2
52 TRIGONOMETRY.
angles equal, but the triangles may have their sides of any length whatever (see fig.) The angles of a triangle are generally designated by A, C, and the sides opposite to them by the letters a, c.
The first proposition which we set out to prove is the fol¬ lowing :
The sines of the angles of a plane triangle are proportional to the sides which subtend them; that is,
sin A : sin B :: a : b. From any angle 0 of the triangle ABC,
let fall the perpendicular CDj then, by the definitions,
CD .. CD —c = sin Ax and sm B,
dividing the former by the latter, we have sin A CD CB CB. shTB ~AC' CB~ AC'
i.e. sin A : sin B :: CB : AC:: a : b.
In a similar way, by letting fall perpendiculars from A on BC, and B on AC, we have
sin B : sin Cubic, sin A : sin C:: a : c.
AVe now proceed to find the cosine of an angle in terms of the sides.
^When the triangle is obtuse angled. By Euclid, Book II. Prop. 12.
B C2=AB2+A C2+ 2AB.AD... (1). 4 7)
Now — =cosCAD=—cosBAC, AC
because the cosine of an angle is equal to minus the cosine of its supplement;
AD = — AC cos BAC,
this substituted in (1) we have BC2 = AB2 +'AC2 — 2 AB. AC cos BAC
TRIGONOMETRY. 53
a- = & + £2 — 25c cos A; <? + £2 _ (C'
••• cosA=z 2J3 "When the triangle is acute angled, as in the first figure, we
have, by Euclid, Book II. Prop. 13, AC2 = AJP + -5C2 - 2 AS.DB...(2).
— = cos.Z?, D3 = -Z?C cos B;
this substituted in (2) gives ACZ = AB2 +^Cr2 — 2AB.BG
or A2 = c2 4" . cos -5; ^ a2 + c2 — Z>2
cos B= —!— 2ac
Kow, by putting^ for a, and a for c, and ^ for B, we have J2 + c3 — a'
cos -d. =
In tbe same manner
,2 cos^= 2J7
ri dr + b" — c~ cos C = —^ 2ab
17. By p. 27, cos A = 1 — 2 sin2 £ l2 + c2 — d2
2 sm2iA = 1 — cos == 1 — 2bc
2 be — I2 — c2 + a2
2bc a2 — (b2 — 2}g + c2) a2—• (b—c)2
2bo 2bc Now, since the difference of the squares of any two quan¬
tities is equal to the product of the sum and difference of the same quantities, we-have
„ . ,, , (a + b — c)(a—b+c) 2 sin2 % A = 2k W-
-Also, by p. 27, cos A = 2 cos2 A — 1; b2 + c1 — d2 2bc + b2 + c2 — dz
.•.2wsr$A = l +cosA = 1 + ^ = Wo
(b + c)2 — a2 (b + c + a)(b c — a) . . 2bc 2bc
54 TEIGONOitKTRr.
Multiply (1) by (2),
4SmHAcos^AJa + l + c){a + l-e}%-b+c)i-a+i + e)-'
Extracting the square root,
2smiAcosiAz= + * + ^ + + <,)(~a +1 + c)'
but 2 sin J A cos J = sin ^ ;
/ \/{(^ + ^ + c)(tf + ^—^ + c)(—^ + i + c)} ••• sin^= We -W*
This is generally put under a different form by putting the perimeter
a, 3 c = 2/^. T?rom this equation subtract 2^,
a -\-b — c — 2S—2e = 2 (S—c).
JFrom the same equation subtract 2b, a —h + c = 2S —21 = 2(8 — 1).
Also from the same subtract 2^, — a -j- b -f- c — 2S —■ 2a — 2 (S —(i)»
•Substituting these values in equation (3), . , V{2^2(^— c)2(S— b)2(S— a)}
BinA= Wc
_ yfiecg —«)(/&—a)(g—g)} 2bc
^{S(S-a)(8-l)(S-c)} = 2 lo W
This is sometimes done in the following manner:—
sinM = (1 —cos J.)(l + cos A) = ^1 —LlL|_ZZf ^
\ ^ 2bc ) (a + c — + b — c)(b + c + it)(b + c— a)
4&V ' which is the same as equation (3).
TRIGOjS'OIIETET. 55
From equation (1),
8ini^= /(g+^-g) (*-*+0) _ /(^-^) (S-c) a V 4^ -V ^ Prom (2),
003^=, /(i±£±^Xl^) =A /s{s-a). 'V be ■ /\/ be 9
... tan 4-4 = ™M=A /{S-l)(S-e)t / - cos^A 'V 8(8—a)
These values being of great importance, we shall collect them for easy reference. j
sm^ = y{5(5-«) (S—i) (5—c)} (»), be
Wj
«•
; «• Ey the same method as above, sin 1? and sin C may be de¬
termined ; but they may be more easily found from equation (4) thus, put JB for A ; a for b and b for and we have
sin B = —'/IS (S—a) (S-i) (S—e)} (e).
Also putting c for A, a for c, and c for a, we have 2_ ab
sil1 G = 4j:^{s<.s-a) (8-1) (s-c)) (/)•
PEACTICAL OBSEIIVATIOXS.
Equation (a) may be used in all cases where A does not approach near to 90°.
Equation (b) may be used in all cases where A approaches near to 90°.
Equation (c) may also be used in all cases where A ap¬ proaches near to 90°.
56 TEIGONOMETEY.
Equation (d) may be used in all cases where A does not approach near to 180°.
18. The area of the triangle A.CB — & 2 nj)
bnt-^= sin^; CD = ACsmA; -4 G
. ^__A.B.ACsmA /^n 2 V
"Where A represents the area of the triangle. "We may also express the area of a triangle in terms of the sides from the above.
By equation (4),
sin^= 7!S(S-a) (S-b) (S-c)}. OC
This substituted in equation (5) gives
</{S(S-a)(S-h) {S-c)}
*=J{S{S-a) (S—l) {S-c)}.
19. "When two sides and their included angle are given, a sin A b sin JB
Add unity to each side of this equation, a , ^ sin ^ a-\-b sin^+sini? /1N
^+1==s"5r5'"i"1; ~ == —W- Subtract unity from each side of the same equation,
a sin A a-b sin A — sin JB b * = sin-#~~ ' ** b sinU W*
Divide (I) by (2) and we have a + b sin A + sin JB a - b sin A — sin B
But by Art. 17, sin A+sin B = 2 sin J (A+B) cos £ (A-B), and sin A - sin B = 2 sin £ (A—B) cos J (A+B);
TKIGONOMETEY. 57
.. a+b 2sin^+-5)eos|(^-i?)_^ w yy^x; J-Ti) a—b~2sm±(A—Ji)<i03l(A+li) iK ' ~K
tan | (A-^B) cot j- G tan^ (-4—B) B)' since tan -J- {A-\-B) = cot J (7.
In logarithms, log tan i (A—B) = log {a—b) — log (a + 3) + log cot i (7.
From f = s|n ®, we have « sin'^
log ^ = log a + log sin C — log sin A ; l may also be found from the formula,
cos C --= —^r——; which may be adapted to logarithms in 2ao
the following manner: c2 = a2 + b2 — 2ab cos C;
but cos 0 = 1 — 2 sin2 ^ C; c2 = a2 + b2 — 2ab (1—2 sin2 % C)
= a2 + b2 — 2ab + 4ab sin2 J C — (a—by + Aab sin2 b C
-= («—J)2 {! + (^a—bf G} '
now since sin2 h G= { C | - ^-hy " ( a-b )
which, being a square, is necessarily a positive quantity, and as this may be of any magnitude whatever, we may put
taii30= C;
then e2 = {a—hf (1 + taa2 6) = {a—If sec2 6; c— (a—b) sec 0,
which is adapted for logarithmic computation, thus: log c = log [a—b) + log sec 0 — 10.
The following is useful in cases where a and b are nearly equal:
c2 = a2 + b2 — 2ab cos C; but cos C=2 cos2-jV C — 1;
d 5
58 TBIGOHOMETBY.
. \ c2 = a2 + 52 — lab (2 cos2 J C — 1^' = rt2 + Z>2 + lab — 4a5 cos2 ^ C
( 4ab cos2 iC\ = («+^)2 - 4a& cos2 i (7= (a+J)211 (a + ^)» \ '
Now ^ CQS7 * ^ is always less than unity,
_ {a+by for V ab is less than J (^ + 5)^, or (a-\-b)'i is greater than 4a5, and since the cosine cannot exceed unity, it is evident that the above is a proper fraction, and we may put
. ' Aab cos2 -J- C sm 6 — (a+bf '
c2= (a+bf (1 - sin2 0) = (a+bf cos2 0, or c = (rt+£) cos0,
log c = kg (rt+3) + log cos 6 — 10; 2 a* b*
since tan 9 = ^ sin J G,
log tan 6 = log 2 + 2 ^Oo a ^ 2 ^ ^ S^n ^ ^
— log( -»)• 2 a* M
Also, since sin^ — ^ ^ cos J C,
+ 21oS^ + 21
— log (fl+5).
log sin 0 = log 2 + ^ log a + ^ log 5 + log cos h G
OX THE AMBIGUOUS CASE.
20. "When two sides of a triangle and the angle opposite one of them are given, there is evidently an ambiguity; for if OA be taken less than BG, but greater than a perpendicular from G on BA or BA produced, and if with _ G as centre and GA as radius, a circle be B
* f-lL? > f^ab or a + b > 2*fab or a + 2ab -f l? > 4«5, or a2 — 2^7; -f X? > 0, or (a—b)" > 0;
for any even power of a quantity is positive.
TBIGONOMETBY. 59
described, it must necessarily cut the line BAA. in two points, while in all other cases it will either fall short of it or touch it only in one point. When A C is less than a sin B, the triangle is impossible : but when AC — a sin By the tri¬ angle is right angled, and there is no ambiguity. And when A C is greater than B (7, there can be only one triangle, and, if it fall between these limits, there is an ambiguity, for there will be two triangles having the same data; thus when the least side A C is opposite the given triangle B, it is evident that either AB C or A!B C may be the triangle required.
21. To find the area of a triangle in the terms of the radius of the inscribed circle, and sides of the triangle.
The triangle A CB is composed of the triangles A OB> AOC, and BOC; 0 being the centre of the inscribed circle.
Now the area of the triangle A B . OB
AOB =
AOC-
BOC-
AC. OK 2 :
BC. OF
cr : — > 2
= "2 ' ar
Hence the area of the whole triangle
AnT>—ar i ^ , cr_a+l+e 2" "2 + 2"=—2""
Since aJr^^rC — ^ ^e semi-perimeter, we may express
the radius in terms of the sides ; putting, as before, A for the area of the triangle,
A A=zrS; r = JS ' •(f);
but A= */{S(S-a) (S—b) (S — c)} by p. 56;
_ '/{£(# — a)(S — b)(S — c)} 8
_ /(S — g) (S—b) (S—c) (2).
60 TRIGONOMETRY.
22. To find the area of the triangle in terns of the radius of the circumscribing triangle and the sides.
By the 20th prop, of the third book c
of Euclid, the angle A. OB is double the " angle A CB.
Draw OD perpendicular to AB, ( which will bisect the angle A OB and the side AB.
Now by p. 56, aV A C. CB sin ACB AC. CB sin A OD ^ "
2 = 2 ba AB ba 1 AB ale ~'2'~OA~'2"2' OA~TR'
We may nqw find the radius of -the circumscribing circle in terms of the sides; for since
. abc m p abc _ ale f ^ A -—, .'.K— {S—b) (S-tf)}"
Multiplying (2) by (3),
A
^ abc abc KV — 777-=- 4>S' 2
By dividing (3) by (2), R abc r 4(S—a) (S—b) (S — c)'
Examples. (1) In a plane triangle if a, b, c be the sides, and A, B, C
the angles, prove that c = b cos A ± y/(a2 — b2 sin2 A).
b2 + c2-a2
Since cos A = ^ > 2bc 2bc cos A = 52+c2 — a2,
c2 — 2bc cos A = a2 — b2, &—%bc cosA+b2 cos2 A =z a2 — &2+£2 cos2 A
— a2 —*(1 — cos2 A) b2,
TEIGOtfOMETRY. 61
(r — c »3 A + ' ci o2 rr or — V2 sin2 -4, c — f) cos A V (or — b2 sin2 A),
c — l cos A ± VO*2 — 52 sin2 ^4).
(2) Pind the area of a square when the difference between * the diagonal and the side = m feet.
Let x = side of the square, then x + m = diagonal, x2 + x2 = (z + m)2, or 2^2 = (a; + m)2,
cc\/2 = cc + wz, «a5\/2 — x = ?;?,
a? (v/2 — 1) — w,
(3) Given the area, angle (7, and a + b; find the sides of the triangle
■« = ^=I=WS + 1)»,
a:2 = m2 (>v/2 + I)2.
sin C = 2A,
Lot « + 5 = w?, « + 2ab + 52 = wi2,
but a + b nz wj ; *
62 TEIGOKOMETRT.
whence c may be found; or c may be found by the formula
<r = {a + 5)2 — \db cos2 (p. 58). • 2t
(4) Given £, the base of an isosceles, triangle, and p the per¬ pendicular from one of the equal angles upon the opposite
side: shew that the area = ^ — 4>/(c2—^2)
By the 13th Prop, of the second book of Euclid
AC2=:c2 + 1?C2 — 2£C.£D, - c but A C = B C;
JBC2 rzz <» + BC* — 2BC.BD;
2 s/ (c2 — j^2)' ~R area = - BC.AT) $6*
4n/(c2— f)'
(5) If 0 be the angle between the diagonals of a parallelo¬ gram whose sides a and b are inclined to each other at an angle a
then tan 0 = sin a, a' — 43
Let DB = a, and AB = 5, then d1 = C*#2 + Oi)2 + 250. OB cos 6», b ,
b-=--O£3 + AO2 — 2AO.O£cos0, /> . a2 — b2^ 2A O. OB cos + 250. OD cos 0 a ^ c
= IB0 (AO + OD) cos 0 = 2B0.AD cos0, area of parallelogram = ab sin a; but AD .ISO sin 0 is also equal the area, hence bj* division
il siqa _ AB.BOsmO 1 oa — 53 _ 2Z'0.A1) coT0- 2 11 '
•. tnn0 = ^™f a1 — 52
(6) In a triangle AB C prove that sin (A — B) __ a2 — Jr
sin C c1 '
TEIGONOMETET. 63
sin (A — JB) sin A cos I? — cos A sin JB sin C sin C
sin A -n sinJB , — —-cosi?——--cos A sm C sm O
a a2 + c1 — b2 b b2 + c* — d' * c' 2ac c' 2bc a2 + c2 — b2 b2 + c2 — a2
2c2 2c2
a2 — b2
c
(7) Prove that the area of a triangle ABC a2 + b2 -f- c2
4 (cot A + cot B + cot C)'
Since cos^ = ^ — and by p. 56, sin A 2bc } J bo
-r, a2 + c2 — b2 • -n 2 A COS B — ——— , sin^rz ; 2ac ' ac
n a2 + b2 — c2 . n 2 A cos C — ■ ' ^ , , sm C = _ 2ab 1 ab
Ey dividing each in the first gronp by the corresponding one in the second, we haye
cot.4 = coti? = fl2+^-y, cot + 4A 4A ' 4A '
.-.by addition,
• cot C'=
j? + V + c-
cot^+cot.s+cot a=
A =
4A a2 + b2 + c2]
4 {cot A + coti? + cot C)'
(8) Erom the top of a monntafo a miles high, the visible' horizon appeared depressed A0; required the diameter (2r) of the earth, and the distance (d) of the horizon.
61 TKIGOXOMETUY.
In the annexed figure C2I = r + a and since-ZQ/T? = CAT is evidently = A,
CM cos A — B C or (r + cos A = r, a cot A
a cos A r = 1 — cos A 1 — cos A sin A 7
(Dividing numerator and denominator by sin A)
a cot A sin A a cot A. 2smi A cos i A 2 sin21A
= a cot A cot £ A ; 2 sin2-A- A
BC , , r , 1 r —-- = cot A, or - — cot A ; d = ——- BM d cot A
a cot A cot ^ A cot A
= a cot | A.
(9) If L — length in miles of an arc of a great circle of the earth, and D the depression in feet of one extremity of it
2 below a tangent to the other, then D = - I? nearly. 3
AC =7920 miles nearly, AB=L, HB = D feet: D
5280 miles,
*.* the triangles AJIB and ACB are similar; HB : L :: L : AC,
B 5280
: B :: L : 7920,
D.^ = /A 5280
D.%=V,
^ 2Z2 1 r. B — nearly. 3 J
(10) Given the perimeter, and the area of the triangle, and the angle A ; find a.
By p. 55, cos A . /S{8-a) * be '
TMGOKOMETRT. 65
2A ^sin^ = 2A, be — sin J?
cos2 tV A — — sin A = —— . 2 sin b A cos b A. 2A 2A
Divide by sin ^Acosi A, and
, or £0 = /S2—z
$2 —^ A cot -J- rtf
cot -l'A = ——or 8a = S2—A cot bA; A
a = ■ 8
(11) Given the area, side c, and A + £; find A — j5. a. ^ ^2 + J2—c2
Since cos C =— 1
2^i 9
a2 + b2 — c2 = 2ab cos (7, a2 + b2 = c2 + lab cos (7,
4A 2tf5 : sin C'
. ^ + 2^5 + 52 = c2 + cos (7 + sin C
o . 4A ^ , 4A = r + -——cos (7- sin C sin G
=£"' -f- 4A [ cot C\/( 1 -f- cot2 C) J j a -J- 5 = VC^2 + 4A{cot C + \/(l + cot2 (7)}].
In the same way a — h=ij[<?+ 4A{cot C— V(1 +cot2 (7))],
a-\~b tan A -\-jB /c2 -f~ 4A{cot C-f- \/(l 4"Cot2 (7) ] a—b~~~tim%A—\/ c2 + 4A {cot (7-—V(1 + cot2 C)}'
iwU A-Ti) — tn.n 1 ( ^ + B) A + 4A{Cot /(^ + cot2 C) j v ;V c3+4A{cotC,+ v/(l+cot:iC')j.
(12) If from one of the angles of a rectangle a perpendi¬ cular be drawn to its diagonal, and from their intersection, lines be drawn perpendicular to the sides containing the opposite angle, then putting P and p for the last perpendi¬ culars, and D for the diagonal, P* + ^ = D
Let the angle JBA G — Q, then AD sin 0 = DE = P;
^ but by (Euc. vi. 8th Cor.) AB. AD = A C2, AD = ;
06 TRIG OXO^TETET.
but AC = AB sin Q ;
An = AB2 sin2 0 AB AB sin2 6;
and since P = AB sin 0, we have B = AB sin3 0 = 1) sin3 0.
In the same way we have p — B cos3 0,
• B3 — I)2 sin2 0, andjj3 = B1 cos2 0. By addition, x
B3 + 2^ = B3 (sin2 0 -f cos2 0) = _D\ (13) If two circles, whose radii are a and b, touch each
other externally, and if Q be the angle contained by the two common tangents to these circles, shew that a •
sin 0 = b)\/ab («+ t>y n
LetAB=x, OB= i:C=a-\-b, CB = BB-E0 = a-b\ BB BC a —I
cos
sm tt 0 — = = — 2 AB EC a + V
**_✓(! -rin-JO) = ^/{l - (^)} = | 2\/ab
TV
sin 6 = 2 sin ^ 0 cos i 0 = ^ \/ai. (a + 4)-
(14) To find the distance between the centres of the inscribed and circumscribed circles, the figure being the triangle AB C.
Let 0 and o be the centres of cir¬ cumscribed and inscribed circles,
Oo = £, OA = it, on = r,
-Ao = —- -j—-7 , sill I Z OAo= OAC—oAc
= | (180° — AOC) — ^A = 90° — _Z> — % A A' = 903 — i? — {90° — A (i?+ C))
TRIG 02?" OilETItY G'T
&=Ao2-\~Ao2—2AoAo . cos OA.0 a«3 O T-ff
= ^3+ -A-r-i - Tj cos i ( ff-1?); 1 sin2 ^ sm% A - v y 7
(32—i22) sin2^A—f-— 2i?r . sin%Acos%(C — B) =r2—2 fir cos J (-Z?+ 6') cos|-((7—J5);
(32—i22) (1 — cosA) — 2r2 — 2Er (cos ^+cos C), Similarly, {82—j52) (1 —cos <7)=2r2 — 2i?r (cos^+cos^)
Subtract (jP—JRr) (cos G — cos A)— — 2Rr . (cos C — cos A),
... 33—i23 = _2i?r; 33=^ — 2i2r; 8 = ^/(-S3 — 2i2r).
(15) If the angles of a triangle be in geometrical pro¬
portion, ratio L the greatest side 2* §
— 2 perimeter X sin. 12° 51 23''^.
Let P = rt-t-i+c, a = greatest side, A+^A+IA = 180° iA = Z, IA = C,
-4-=102° 51'25"! 7
5= 51° 25' 42"^,
(7= 25° 42' 51" 7
r, , i , (. . b c) ( sin^ , sin C) P=«+8+.=. J1+J+; j = . j l+J^+50 j
_ sin^+sini>+sin C ^ a sin^l 9
I* sin A ^~~sin A+sin i?4-sin G 2P sin %A cos ^A __ P sin ^A
4 cos ^A cos cos JG~~ 2 cos kB voshG*
68 TRIGONOMETHY.
But =£, iJB= a; Psin^ 2P sin cos P sin C cos C
a "" 2 cos C cos J G 2 cos C cos + 0 cos C cos h C jPsinO 2Psin^(7cos J(7 ^
= r-77= r/T" —— 2P sm O, cos-kO cos iu " ' §
=2 perimeter X sin 1^° 51' 25" y. Answer.
(16) The sides of a triangle are in arithmetical proportion, 3
its area=g of the area of an equilateral triangle^ having the
same perimeter, shew that the greatest angle=120°, and that the sides are as 8 : 5 : 7.
Let a — #, a, rt-K^=the sides, then 3«=perimeter;
a = side of equilateral A, and ~ = half its base, 2*
// o «2\ /4a2—rt\/3 .. .... _ a V _ij^ V 4 = IT" altitudes^ V 3.
f1
iN'ow both A's may have the common base a; area equilateral A : area of the other :: altitude of equilateral A : altitude of the other;
or> 1 : 5 :: 2n/3 : Yo^3 =^-0,
=«2+2ay/{{a — «)8—^a2 j +(« — xf-,
/ 27 or, 2ax = a2 + 2a ,* / {(a — x)2 — a2 j — 2n.r, •
4a« = «2+2rt/Y/ {(« — cc)2— ^«2}.
TBIGONOMETHr. 69
27 162?— Stfcc + «2=: 4 (a — ^)2 — —a2
2o
— 4a?— Sax 4- 4a? — — a*. ^ 25 '
27 12 a:2 = 3a? ——a?,
J/O
4«-
25 '
, 4(j2 £ a;- = — : x == : rt, 25 ' ts ' 3
« — x = o
, 7 a + ^ = -0,
AC=Z, £0=5, AB = 7, 3^ >7(i . ^
— are the three sides, which are as 3«, 5tf, 7^, 5 5
or as 8, 5, 7 ; _^a+i?^2 —^2_32+ 52-72_ 15^ 1
COS C —- 2A0.BG ~ 2x3x5" 30— 2'
but cos 120° = — i, therefore C = 120°.
(17) The angles of a triangle are as the numbers, 1, 2, 3 ; and the perpendicular from the greatest angle upon the
2«2
opposite side is p. Shew that the area = —, V3
A + 2A + 3A = 180° ; A =2 30°; 13= 60°; C= 90°.
sin A : sin A CD :: p : AD, (fig. p. 52)
70 TEIGOXO^EETKY,
AD = p sin A CD sin A
V3 9
2 sin B : sin i?CZ) :: p :
* 1
2
(18) The angles of a triangle are as tlio numbers, 2, 3, and the side opposite the greatest angle =100 feet; find remaining sides and the angles:
log b = log c + log sin B — log sin (7, log c = log 100 = 2#00000
log sin D = log sin 54° = 9*90796
11-90796 log sin (7 = log sin 90° = lO'OOOOO
1-90796 = log 80-9. sin C : sin A :: c : a}
2A + 3 A + oA = 180°; A=:S60)
C=i 90° ) sin C : sin B :: c : b,
7 sin B o = c -—
TKIGOXOMETKY. 71
log a = log c + leg sin A — log sin C log c = log 100 = 2 00000
log sin A — lop: sin 36° = 9-70922
11-76922 log sin C = log sin 90° = lO'OOOOO
* 1-76922 = log 58-78
a = 58*78 ) • b = 80"9 jsll'es-
(19) Find a point within a triangle at which the three sides shall subtend equal angles :
BC — a\ AP — x AC=b\ BP = y AB = c) PC=%
Then z APB = BBC — ABC = 120°;
the cosine of each = — and the sine of each — ; 2' 2 '
a~ = BB- BC- — 2i?P . BG cos 120°
= BB- + BC2 - 2BB . BC ^
= BB- + PC2 + BB . BC, b2 = AB2 -f BC2 + AB . BC, c2 = AB2 + BB2 + AB . BB a2 + I2 4- c2 = 2x2 + czy2 + 2z2 + xy + yz + xz.
Let A AB C = in2, m2 — ABBC + AABB + AABC-,
_ yz sin 120° | xy sin 120° l xz sin 120° .. m — - 1 - 1 -
= -j- (xi/ + yz + xz)>
4m2
xy + yz + ^ = y3»
Qxy -J- Sys m 4??^" \/ 3. But 2 (x2 + y2 + s2) + {xy + yz + xz) — a- + b2 + c2; 2 (x2 + y2 + z2) + 4 (xy -\-yz-\- xz) = a2 4- ^ + c2 + 4m2 */3,
A
72 THIGOXOilETRY.
+ y"' + s2 + 2(^y + y.~ + scz)—^- ^ e + 2m- ^/S^,
x+y + z=x/+ 2m, ^ = p>
1° + y + a = p, x + y=P — z>
x~ + 2xy + = (y3 — a)2, but x* + y* + xy = c- ; xy=z(p — zy — (?.
Also, sinco + ara + yz = —
4M2
« (y + ®) 5 tut as + y + « = p ;
4wi2
4m2
•'• O — «)* — e2 + « (P — «)= ^3 :
/ N / N O ^m2 or (p — z + g) (p—%) — <? — ;
o «, 4 m2
F—pz — f=-jg
f 8 = ^ ^ - 7i = " + 2 + C°'+2w 3 -c2 - 75 __a2 + J2 —(!2 + 2»t2_
2 ^3' 1 /a2 + F- — c2 2wi2\
•'• z--p\ 2 +7^} i /y1 + c2 — a2 2w2\
Aiso x — — ^ ^ +—),
1 /a- + c- — I- 2»}2\ "ia!'=^V 8 + —s)-
(20) An object 6 feet liigl), placed on the top of a tower, subtends an angle = tan-1 *015, at a place whose horizontal distance from the foot of the tower is 100 feet, determine tho - tower's height.
TKIGOXOMETEY. 73
Let BAB = a, AB — x,' AB = y, then tan a = *015, a = 51' 83" ; sin a = 'OISIS, cos a = -99989 nearly.
2 A BAB = xy sin a = -OISIS xy. Also 2 A ABB = A C x BB — 100 x 6 = 600 ;
.\ 'OlS xy — 600, xy = 4000
»2 + y2 - 62 „„ ^ + y% - o3 COS a =
•9999 =
2xy x~ y~ — 36
80000
or '9939 2xy
x2 -j- y2 = 80028
2xy = 80000
x2 -j- 2xy -f- y~ = 160028 x2 — 2xy + y2 = 28 x + y = 400,035 x — y = 5-291
2a3 = 405,326 tc = 202-663
2y = 394-744 y= 197*372,
BC—^Jiy1- AC2) = y/(l97-3722 - 1002) = ^28955-76384 = 170-23188 feet.
(21) Prom the top of a mountain 3 miles high, the -visi¬ ble horizon appeared depressed 2° 13' 27"; required the diameter of the earth, and the distance (d) of the horizon.
Referring to the figure and formulae at page 64, tyc have r = a cot A cot £ A,
r — = cot A, r=d cot A;
a cot A cot i A = d (fot A d= a cot A.
74 TRIGOXOHETKY.
Here a = 3 miles, and ^ = 2° 13' 27"; .\<? = Scot log 3 = 0-47712
log cot 1° 6' 43*"= 11-71194
12-18906 10
2-18906 =log of 154-54 ; .'.<? = 154-54,
r — acoiA cot\ A — a cot 2° 13' 27" cot 1° 6' 431",
= log a + log cot 2° 13' 27" + log 1° 6' 43|" 2< log 3 — -47712
log cot 2° 13' 27"= 11-41075 log cot 106'43J" = 11-71194
(22) A person, at a known distance from two towers, observes that their apparent altitudes are the same; he then walks a given distance towards them, till the angle of eleva¬ tion of one is double that of the other; find the heights of the towers.
Let x and y = the heights, AB = a, and a, (3, the angles,
23-59981 20
3-59981 = log 3979 .'.r= 3979; 2r = 7958.
AB = b, CB = c, CD = d,
tan a = — = 7-, tan j3 = ~ , tan 2/3 = — ; a b' d* c
but tan 2/3 ^ 1 — tan3/:* 2_ 2r d2 — y2 c '
2 tan /3 d __ 2 a;
TKIGON 03IETET. 75
and x — -p, b
al 2d)j b ay -d a
d' — tf' c Ic' cP—y" Ic'' 0 Idcb
{p— y~) a — Sflco ; d-— y- = ,
, 2dcb
-s/i*-™)-
■=?-V(^") // 2c5\
X= T "V ( ~ "ad/ (23) In the arc AB of a circle, centre 0, AO is taken
= sin AB; then will sector COB = segment A OB. The sector COB -j- sector COA = Ai? OA -j- segment B OA,
or sector C OB -{-AC — 2
= segment A CB + BJ) ,
hut BB = arc A C; sector COB= segment ACB.
rnACTICAL EEMAEKS OX THE SOLUTION' OF TRIANGLES.
23. In -working out the various problems in the solu¬ tion of triangles by the tables, when one or more logarithms have to be subtracted, it is better to write down their com¬ plements, or what each of them wants of 10, instead of the logarithms themselves, and then add them altogether, sub¬ tracting as many tens from the index of the sum as there are logarithms to be subtracted.
e 2
76 TBIGONOMETEY.
Thus, if the logarithm to he subtracted he 3*6947362, it is the same thing as to add its complement 6*3052638, and then to subtract 10 from the sum; the easiest way to per¬ form the subtraction is to begin at the left hand, and sub¬ tract each figure from 9 except the last figure on the right, which must be subtracted from 10.
If the index of the logarithm be greater than 10, write down whaHt wants of 19, and take the rest of the figures from 9 as before; observing, that after the addition is made 20 must be taken from the index. Thus, the complement of the logarithm of 13*6874563 is 6*3125437. If the logarithm of a decimal which has a negative index is to be subtracted, add the index, considered as positive, to 9, and then find the complement of the rest of the figures as usual, and subtract, 10 from the index of the sum. Thus, the complement of the logarithm of 2*3674586 is 11,6325414.
24. Before proceeding to the next examples we shall shew how the trigonometrical tables are constructed. (See Lefebure de Fourcy's Trigonometrie, p. 35.)
Let us at first find the sin 10"; we have before stated that if the diameter of a circle be unity the circumference = 3*141592653589793 = tt, and when the radius is unity the semicircumference = tt, and since there are 64800 seconds in 180°, we shall have in parts of the radius
arc 10" = —~— = -000048481368110. 64800
Now in very small arcs the sine and the arc are very nearly equal, hence the above may be considered as a good approximate value of the sin 10".
25. "We proceed now to shew that in the first quadrant, the arc is greater than the sine and less than the tangent
Let AP be the sine of the arc AH, and ^^Tthe tangent, and suppose this figure to turn round OP till the point A falls on C, we shall have the arc AO> the chord AC, and, consequently, the arc ABy> AP; hence, the arc is greater than the sine. We have, also, the arc A G < (AT+ CT), and therefore AB < AT
Hence it follows that if differs little from 1, the sin a
ratio t—— differs still less, sm a
TRIGONOMETRY. 77
Also, as the arc decreases the ratio of that arc to its sine becomes as near to unity as we please; thit is, this ratio has unity for its limit.
sin a tan a 1 And since tan a — ; .*• ^ • cos a sm a cos a 0 A
Now, diminishing the arc (which is supposed less than 90°), the cosine increases and approaches unity as nearly as we
please: that is, the ratio —— or its equal ^—- continually cos a sin a
diminishes, and has unity for its limit. The arc being greater than the sine and less than the
tangent, the ratio -r— can never be less than l nor greater sm a
tan a . , than ——, and since this last ratio can approach unity as
sin a nearly as we please, the former one will also approach unity as nearly as we please, and it is on this account that we take the arc of 10" for the sin 10".
26. "We must now proceed to determine the degree of approximation
sin a =« 2 sin ^ a cos ^ a, and tan ^ a > ^;
Sin 2" 1 .11 1 or p > - .-. sm > - a cos cos - a
. 1 1 or 2 sm - a > a cos - a; Ji A
1 o • 1 O 1 • ^ ,1 .and 2 sm - a cos - a > a cos' - a, or sm a > a cos- - a;
but cos2 - a — 1 — sin2 - a ; 2 2 '
sin a > a ^1 — sin2
• ^ (, Z1 V i ^ «3 sm a > a j 1 — »>«_-.
78 TEIG0N02IETIIT.
Applying this result to the arc of 10", taking the nearest value to five places of decimals, we shall have the arc of
10" < -'00005, then ^ (arc 10")3 < -000000000000032;
hence we have sin 10" > {-000048481368110 — -000000000000032};
or sin 10" > *000048481368078,
we see that this sine only begins to differ from the arc of 10" at the 13th place of decimals, and taking the nearest decimal it is the same; hence it follows that we may take
sin 10" = *0000484813681,
and we may be assured that the error will be less than unity -at the 13th place of decimals,
and since cos 10" = ^1 — sin2 10", we have cos 10"=-9999999988248.
27. AYe can now successively find the sines and cosines of 20", 30," . . . and so on up to 45°, by means of the formulae (1) and (4), given at page 28,
sin (a + 3) = 2 cos b sin a — sin (a — 5), cos {a + V) = 2 cos b cos a — cos (a — V) \
we can consider the arcs a — 5, tf, a + £, as three consecutive terms of an arithmetical progression whose common dif¬ ference is £, if we put t, f for these three terms, we have
sin {' = 2 cos b sin f — sin t, cos t" = 2 cos b cos t* — cos t;
• consequently, to obtain the sine and cosine for every 10" we must make b = 10",' and putting a and /3 for the known values of the sine and cosine of 10", we have
sin 0=0, sin 10"=: a, sin 20" = 2/3 sin 10", sin 30"= 2/3 sin 20" — sin 10", sin 40" = 2)3 sin 30"— sin 20",
&c. &c.
cos 0=1, cos 10"= /3, cos 20" =2^ cos 10"— 1," cos 30"= 2p cos 20"—cos lO'7, cos 407 = 2/3 cos 30"—cos 20",
• &c. &c.
TKIGONOMETEY. 79
But as p differs very little from unity, 2/? is nearly = 2, and the calculation can be still more abridged; put 7c for the difference 2 — 2/3, we shall have h = *0000000023504, and 2$ = 2 — Icj consequently the value of sin f becomes
sin t" = 2 sin t' — h sin l! — sin t; sin f — sin t* = (sin V — sin t) — h sin f.
In such a long series of calculation the errors multiply considerably, and we therefore cannot be sure of the decimals being correct to the end, but, by using the formula) given in Art. 15, we may verify the results, for the number of deci¬ mals common to both processes may be relied on as exact.
"We may remark that when the radius is unity the sines and cosines will be fractional, and, consequently, their loga¬ rithms will be negative, in order to render them positive, the radius is taken = 1010.
EXAMPLES.
In the right-angled triangle A CB, given the liypothenuse AB = 1785*395 yards and the angle ABC = 59° 37' 42", to find the other two sides.
To find A G or h. 1= 0 sin H,
or log S= log c + log sin £—10 • log 1785-395 ...3-2517343
log59° 37'42" ...9-9358919
To find jBC. a=:c cos JB,
log a = log o + log cos £ —10 log 1785-395 . 3-2517343 log cos 59° 37'42" ...9-7038132
log 5 3-1876262 i = 1540-0374.
log (t 2-9555475 .-. «= 902-708.
In the triangle ABC (see fig. p. 52), given a = 9459'31 yards, b = 8032*29 yards, c = 8242*58 yards, find A;
2jS=25734,18, 5=12867-09, £-£=4834-8, 5-0=4624-51,
log sin k-A=10 + i {log (5-5) + log (5—c)—log I — log c],
2 log sin J A = 20 4- log (S—V) + log (S — c) — log S—log 0,
80 TBIGONOMETEtf.
log (S — b) = 3*6843785 log(iS—c) = 3*6650657
Comp. log b = 6*0951606 Comp. log c = 6*0839368
21ogsiiii^ = 19*5285416, log sin 1-^ = 9*7642708,
IrA = 3503r47", " A = 71° 3' 34".
CHAPTER IT.
ON POLYGONS.
28. To find the radius of a circle described about a regular polygon of n sides.
Let It = A C = radius, AB one of the sides of the polygon — a, A CB the angle subtending one of the sides of the polygon, then n times the' angle ACB -will be the sum of all the angles, or 360°;
i.ACB= —; z_ A CD = — n n
since CD bisects the angle A CB, R sin A CB -= AD, or 2R sin A CD = AB;
AB AB 1 ^ 180° •*. It = 7; :—7777: " To~u — o cosec 2 sm A CD 0 . 18 2- n jt sm ——
n 1 180°
. = - a cosec . 2 n
29. If now we suppose AB to be the side of a regular polygon described about a circle, then CD will be the radius of that circle, which call r,
a 00 <• 1800 then -r- = cot A CD = cot : AD n
'JKIGOXOJIEIIIY. 81
180° 1 , 180° 1 ,180° CD = r =ADcot -=^ABcot—=^a cot —;
„ - 1 / 180° , 180o\ R + r = 2rt(cot— + coscc —)
180° (cos )
1 \ 1. . 7. < i vt\o ' <> 1 . 180° • 180° f - / sin sin ) v. n n J r. , 180° at 90o\
_i_(1 + cos—) ,90° 2 ^ / . 180° f 9(F 90° 2 n '
V Sm~ J 2sm —eos — ,180°
1 180° 180° 1 3cosec" n it . r — t or cosec cot = ia i 4 n n 4 isqq 7
sec n 1 180= 180°
7. ^ «cosec — cosec —— It 2 • n n 1 r 1 180° ~~ 180° 180° 180°
o « cot cos —- cosec —— cos ——
30. To find the area of a regular polygon of w sides in¬ scribed in a circle.
The polygon is composed of as many triangles as there are sides of the figure :
. 180° 180° area of the triangle ACB = AD. CD = r sin—jj— r cos-^—
180° 180° 1 0 . 360° = r2sin cos = - r-sm n n 2 n
Hence the area of the inscribed polygon = n times the tri-
n+i 4V. «. * . . 1 a my AC. CD sin A CD Or thus, the area of the triangle A CD = ^
0 . 360° sm n
i a m~> n - • 3600 angle A CD = p; r~ sm 0 In
82 TBIGOXOMETEY.
n 360° area of polygon = 5 r1 sin' Ji
31. For the circumscribing polygon. AT) 1 R0o
Let CD = r, 77^ — tan ACD = tan —; ' CD n
n-m 1800 * 1800 . \ AD = CD tan — r tan —-, n n
180° ^ area of A ACB = AD . Ci) = r2tan , A
n
area of polygon = n times the triangle A CB — nr2 tan 180°
n j . 360° 1 . 360° area of the inscribed polygon 2 ^ sin 2-Sin n~
area of the circumscribed polygon — 180° — 180° nr2 tan tan
1 . 180° 180c
o • 2 sm cos 2 n 11 180°
180° ■=cos2 n
180° cos -
The area of a circle whose radius is r = tt r2.
Prom the above the area of a polygon inscribed in a circle, is
2 TT n . . 360° 11 „ . 27r o t" sm = r. r sm — == ?rr~ 2 n 2 n
As we increase the number of sides, the area of the polygon approaches nearer and nearer to that of the circle, and when n is infinitely great it becomes equal to it, for then
2 TT 2 w
sm ■ — is an infinitely small angle, therefore ■■ n 27r
area of the circle = Trr*. n = i;
TEIGONOMETRY. 83
Examples on the two preceding Chapters.
(1) Given the perimeter of a right-angled triangle, and the radius of the inscribed circle ; determine the sides,
s=a + y±c) ^ = A = 8r,
(11 = 2 Sr, ^^-j-S + ^! = 2/S,, 4 + J = 2& — c, a+h = 2S—e=2S-(S-r) = 8 + r,
a + b = 8 + r, c? + 2 ab + 52 = /Sf2 + 2 Sr + r\
4ab ' 8 Sr; a? — 2 ab + V = S* — 6Sr + r\
a-l^^St — eSr + r-) (i b : 28 — Cj
2a = 2S—e + 82 — &8r + r2), _ 2S - e + V(<S2 - SSr + r2) a ,
2b = 28 — o — V(82 — 6Sr + r2), p _ 28 - c — V(S2 - .68r + O,
2 o = S — r + 2 + i2 = 4S2 — 4^ + c2
a2 + V = c2
2 ab = 4>S2 — 4/Sk, but ab = 2$r, 4iSr ^ 4/S2 — 4j& 4^ = 4^2 ~ 4^r
4 c = 4/S^ — 4r c — 8 — r.
(2) Given the perimeter of the same triangle, and the radius of the circumscribed circle; find the sides,
a + b + c _ ^ ab abo 2 T A "" 4i2 >
^ 5 -}- c 2S 2R =■ t?, «+ b = 2S- 2X,
84 TEIGONOMETEr.
a2 + lab + J2 = 4/S2 — 8Si2 + 4^3, a3 + J2 — 472";
2rt5 =4S2-85,i?; , (C-— 2a5 + 42=4i23 + 8/SiJ-4-S2;
.•.a-l = V(4iJ3 + 8^i2 — 45s), « + J = 25 — 2iJ,
2fl = 25 — 2R + v'(4i22 + 85i2 - 452), 25- 2i2 + V(4-ffi- + 85iJ — 452)
« 2 '
25 = 25 — 2iJ — ,v/(4iP + 855 - 451), , 25 — 25 — ■v/(452 + 855 - 45s) 5 = 2 '
(3) If (JD) and (d) be the diameters of the two circles, and (0), (&) the sides which include the right angle; shew that 2) -f- d = ci -f* b.
AF+FC+ CD+DB^b + a, AF= AFand BE = BD, ^ FO=r andi)C=r; ^ ^
*. * AJS -j- BF 2r cl -4™ b, but AH + JE£ = D. A ^ "
Hence a + b = l)+2r = l) + d.
(4) The sides of a triangle are as 3, 5, 6 ; shew that £ : r :: 45 : 16,
p abc 3.5.6 __ 3.5.3 45 ir = 2(a + i + c) "" 2 (3 + 5 + 6)"" 3 + 5+6 "" 13'
= ^/(5-a)(5-g)(5-0)_ ^/(7-3)(r-5)(7-6)_^/8
„ 45 45 . /8 45 /7 14"5"r—14 ' V 7 14 /\/ 8'
45 /7 /8 It: r-.: ^ X ^,
45 X 7 : 14 X 8; Jt: r :: 45 : 16.
TRIGONOMETRY. 85
(5) If the side of a pentagon inscribed in a circle be 1, \/(5+N/5) the radius = — .
Here 11=5, a = l,
perimeter = 5 = sin- n
n r . 180° = 2 x 5 r sm— o
l=:2r sin 36°
= 2f
r =
_ or \/(5— s/o) 2^2
— r \/(5— \/o) 72 '
1 V (5 — \/5)
n/2
= y_2 n/(5-N/5)
\/2n/(5 + \/5) 7(25 —5) 7
\/2\/(5-f V 5) ~~ ^20
\/(5+n/ 5) V10
(6) Pind the actual value of a side of a twenty-four sided regular figure inscribed in a given circle,
15° Hero » = 24. Put A=—, 2 A = 15°,
n/3 — 1
86 TKIGOXOilETIir.
Perimeter = 2«r sin IT n
180° = 2>^24r sin
24
= 2x24r sin —, 2 '
but sin A = 1 { V (1 +sin 2A) - V (1 ^— sin 2A)}
-°[r {V^+J^)~\/('~yh^)i
but a — Per™e':ei' u a 24 '
a = 24r 24
(7) If r be the radius of the inscribed circle, area = r1 (cot J^+cot J^+cot J C),
By Euc. Eook IY. Prop. 4, Zl OB A =%£, ^LOCA = iC.
Now J3a = Oa cot J i?, Ca = Oa cot J C\ a = Oa cot £.#+ cot J (7, a = r (cot-J-i> + cot b C), J = r (cot A A-^cot J 67), #
c = r (cot J A+cot £ 2?), fl-j-j-j-e (cot Ji?+cot-J- C)
+r (cot J-i+cot J C)+r (cot i^i+coti^),
TJilG OXOMETEY. . 8 7
Q^rCcotJjfr + cotiC) + r(cot:^A-\-cotiC)-rrKcotiA + cot^B) 2
__ (2 cot iJ- + 2 cot £ J?+2 cot | C) ~~r 2 _ 0 . (coti-i+cotJ^+cot-J C)
2 ' /Sf = r(cotiA + cot Jj^+cotJ 0).
But area = S. r, area = rr (cot-J-^+cot^^+cot J C); i
area = r2 (cot -l A + cot + cot iC). (8) The perimeters of an equilateral triangle, a square,
and a hexagon, each including the same area, are as ^27,
Let each side of the AAJBC at p. 82 = a, 2a = perimeter,
GD =y/a1 - aild ^ = areaA;
side of equal square = ^
=/^VlsVfVi-v/il''
4 . ^ Perimeter of square.
Area of A = CD. |=
= area of square = area of hexagon. Let x = side of hexagon. 4
, n, nx2 , tt Gx2 ,180° 6a;3 , OAO aVrca of hexagon = —. cot - = — . cot —— = — cot 30° 0 4 n 4 6 4 3 2 /Q /3a4 27s4 3a4 ,
= 2 \/16' 4 = 16' =
Ck3 = a2, cb -v/6 = «; .*.:« = = side of hexagon: a/G
-!L = a \^6 = a \/3 . ^2 — perimeter of hexagon: 6^/2
3(1: 2a^3 : aV2.i/3= ^27 : ^16 : ^12.
88 TRIG OX03IETRY.
(9) If 0 be the centre of the circle inscribed in the triangle ABC,
0*A.~ OB2 -j™ OC~ — (ib cic be — 12J?r. See fig. page 59.
AO2 = r2 + AE2, = t* + r2 cot2 *) OB~ = r3 + BF2 — r- + r~ cot2 \B, C 00* = r2 + ^C2 = r2 + 7'2 cot2 ^ C, J
A0"- + OG- — 3r- + r2 ^ cot2 ^ + cot2^ + to?^,
by Art. 21, and p. 55, _ 3 (s - a) (.8 -V){s- e) (i - a) (a - V) (s - c)
s s f 8 (s — a) s(s- b) s(s — c) } {(s — 3) (5 — c) (5 — a) (5 — 0) (5 — &) (5 — a)J
_ g (« -«)(«- &) (« - 0 + (s _ ay + ($_ by + (s _ cy S «
= 3.s2— Ssa— Ssb + Sab — 3sc + Sac + Sbc— s
+ 6'2 — 2sa + or + s2 — 25^ + 52 + s2 — 2sc + c2
— 6s2 — 55 (rt + b + c) + a2 + i2 + + 2tf3 + 2^
+ 2^ + ab + ac + be — 5
= 6s2 — 55 . 25 + (a + b + c)~ + ab + ae + be —- 12i?r = 6s3 — 10s2 + 4s2 + ab + m + be — I2fir ~ ab + ac + be — 12itr;
.\A02+ OB2+ OC2 = ab + ac+bc—12rir. (10) The area of a regular hexagon inscribed in a circle
is a mean proportional between the areas of the inscribed and circumscribed equilateral triangles.
Area of hexagon = sin — 2 n 6r . 3603
= — sm 2 6
= 3r sin 60° 0n/3
TRIG ONOM KTIiY'. 8 9
Area of inscribed triangle = ^- sin — 0 2 n 3r2 . 360°
=Tsm-T 3r2 .
= sm 120°
VS — 2 2
3r2 a/3
Area of circumscribed A = nr2 tan - 2
= 3 r2 tan 60° = 3r2 ^3;
3rV3 3r2V3 3r2 v/3 00 /0 as : —5*- :: —: 3r t/3.
(11) The square of the side of a pentagon inscribed in a circle is equal to the sum of the squares of the sides of a regular hexagon and decagon inscribed in the same circle.
Pentagon, perimeter = n x = 2 nr sin - ; n
. 180° AB = 2r sm = 2r sm 36° 5
„ V(o — V5) ' V(5 — V5) = 2r 2V2 =r 72"'
Hexagon, perimeter = n = 2 wr sin
. 180° AB= 2r sm = 2 r sm 30° 1 6
= 2r^ — r; AB?=^r\
90 TEIG0X03IEXRT.
^ .180 Decagon, perimeter = n = 2 ?;r sin
u 180°
AD.. = 2r sin = 2r sin 18° " 10
\/ 5 1 Z^/O 1 - 2r = r
4 2
ABf+AB,;- = (3-y5)f' + ,■» ^ (3-^+2) r
(12) Three equal circles touch, each other; shew that the space between them is nearly equal to the square described upon a fifth part of the diameter; find the area when the circles are unequal, and the radii as the numbers 1, 2, 3.
Let rad. = — diam. = 1, tt ■= 3'1416, Ji 2(3-1416)
arc mn = g = o26b x 2
= 1-0472. 1
Sector Amn = ^ r X arc mn = -5236;
the area of the three sectors = 1-5708, Cn2=AC2—An2; Cn = */(A C2 — An2)
= y(AC2— 1)= ^(4 — 1) = V3;
[\ABC = AB* Cn = 1-73205.
Now the space mno — L\ABG— 3 sector Amn = 1-73205— 1-5708 = -161.
1 . 4 Now the square described on - of the diam. = — — -10. o 2o
/diamA2
Hence space vino = (—-— J nearly.
iraGONOHEinr. 91
(13) Let rad. Co — 1, Bo — 2, Am — 3, then the diameters are 2, 4, 6 ; and the sides of the A AJB C are
BC = 3, AC= 4, AB — 5, AB2 = AC* + BC*; BG
zl C — 90°, sin a — AB'
logsmA=logBCf—log .<4.5+ 10, log 3 = -47712 log 5 = *69897
10 log sin 36° 52' = 9-7781^;
A .= 360-8666,
360° : 360,866 :
sector Amn
360° : 530'133 :
sector B?2o
B = 530,1333;
3-1416 x 6 : 1-9305 = arc mn, 1-9305 x 3
= 2-8958, A
3*1416 x 4 : 1-8546 = arc no,
1-8546 X 2 = 1-8546,
circumference of © moC 3'1416 x 2
sector Com 1-5708 x 1
2
2-8958 1-8546 •7854
= 1*5708 = arc mo.
- *7854,
sum of sectors = 5*5358
{\AB C = AC X -DC=t^l~ Q .
2 2 ' . space 7wio = AABC — sum of sectors =*4642.
(14) Given AB = a, BC = I, CB .= c, three sides of a quadrilateral, and ^ ABC= 0, ^BCD = 0; shew that twice the area =al sin 0 + cb sin 0 — ac (0 + 0).
92 TRIGOXOMETEY.
Let AB CD be the quadrilateral. From the point A draw AE WHO,
and froin the point I) draw DF || to AB or EC,
then cb sin 0 = 2^13 CD, and al sin 0 = the figure AB CE
= UABFG + GFCE, = 2&.ABD + □ GFCE-,
al sin 0 + lc sin 0 = 2&.(ABD + DCD) + □ GFCE = 2 quadri. + □ GFCE = 2 quadri. + 2 A DEC = 2 quadri. -f- ac sin (0 + 0) ;
al sin 6 -f- lc sin 0 — ac sin (0 + 0) = 2 quadrilateral.
(15) If R and r be the radii of the circumscribed and inscribed circles of a triangle :
r = 4i2 sin %A sin bB sin C. See Ex. p. 86.
-h{a 1 + c) = r {cot %A + cot |B + cot ^ C), area = r2(cot %A + cot bB + cot \ C), but $A + bB + JC = 90° ;
, by Ex. 26, p. 45, area = r2{cot %A + cot hB + cot bC i
= r~ cot %A cot bB cot C. And by Art. 22, Chap. III., and formulae, p. 55,
area = 2JZ22 sinyl sini? sin C = 4i22 sin bA cos &A 2 sin ^B cos iB 2 sin ^C cos bC = 16722 sin \ A cos \A sin bB cos ^B sin hC cos C = r cot %A cot -IB cot i-C
0 cos £A cos \B cos \ C ^ r sin bA sin hB sin iC
r-. \ ^ 7> ttt— 16^2 sin $A sin sin ^ C, sm IrA sm hB sm £ 6 ^ ^ 2
TRIGONOMETEY. 93
r2 = IGi^2 (sin iA sin |Z? sin bC)2; r — AR sin |A sin sin %C.
(16) In the triangle ABC take a point D, join DA, DB, DC) given AB, AC, the /_ABD, Z.ACI), and Z.BBC, find B C.
A AB — a, /_ABD = a,
d[ \ A C — b, Z-A CD - Z.BBC — j,
u ZBBC+ zDCB^tt — r, ^ABC Z A CB = (tt — y + c -f" /3), Z. BA C — tt — (tt — 74-0+ fi), Z BA C = y —a — /3,
BC = ^(AB1 + AC2 — 2AB.AC cos BAC), BC = yj{a1 + &2 — 2aI cos (y — a — /3)].
(17) ABCB is the circumference of a p
circle, 0 its centre, from C draw a tangent to the circle meeting the radius OA produced, in P, join PB-, then if CP—a, BP=l, 6 = angle formed by BP and a tangent at I),
ai p prove that r = —cosec 6, r being the
radius of the circle. COP is a right-angled triangle ;
OP2 = a2 + r2, and in A ODP, the sides OB,BP, and Z. ODP are known.
OP2 = b2 + r2 — 2lr cos ^ + O^j ;
a2 4- r2 = b2 -f- r2 — 2br cos 4- 0^ ;
a2— b2 — Ibr sin 0; ; rt2_J2 fi2_i2
^ ^ = -r-,— cosec 0. 25sin6' 23 (18) Given the three perpendiculars from the angles on
tl.e opposite sides ; find the sides.
' ^4 TEIGOXOHEIEr.
Let AB = x, AD = y, I3D = z, AC=a, £E = \ I)F=c, 2A = cx = by = az,
2A = 2V{S(S — X)(S — !/)(S — Z)};uL C U
w = 2V{fif(5 —a;)(iSf —y)(5 —a)}
_ o //^ + y + g y + g — ® K + « — y « + y—z\ V \ 2 2*2 2 )
-W\ (*+"+7) (j+f-) («+"-!)(«+?- f) i
-IViO+WJG+i-OK:-^1^-;)}.
2,,=a,\/|(i+j+j)(l+«_i)(i+s""l)(i+?_0!i
2c
V7 {(1+?+5)G+5_1)(1+5_l)(1+S-;)} Similarly, 2b
-\/ {(I+^+i) (1+«~0} 2a
(19) If ^ = the base of a polygon, abc, Sea., the sides beginning from the base, a/3y, &c., the angles made by alc}
v &c., with the base, then I = a cos a + ^ cos ft + e cos y + &c-
Let = I, from the upper angles FiF-pi.\. draw lines perpendicular and parallel to -—■*» the base. Then (fig. 1), ,f ^
Am = a cos a, »m = I cos ft, /' j \ J?0 = C COS 7, oB — d COS 2, A TO n "o K
= yl?;j + mn + no+0 B = a cos a -f b cos ft+c cos y + d cos $; I = a cos « + i cos ft -f c cos y + &c.
fKIGOXOlTETRY. 95
T?' r. 11 r \ .,:Fl!=-2- Fig 1; — = COS (a — w) = — COS a a
— Am— a cos a, mn = Am± An — I cosJS no — c cos y, oB — d cos
, AB — —Am 4- Am + An + no + oi? m A n
— a cos a -f- b cos /3 + c cos ft + &c.; I = a cos a + 5 cos ft 4" cos y + &c.
32. The two following rules may be found useful to practical men ; the first is when two sides and the included angle are given, and the second when the three sides are given.
-n . , a + b tani(A + £) By Art. 19, a — b — tan ^{A — £)'
Hence the following rule : As the sum of any two sides of a plane triangle is to their
difference, so is the tangent of half the sum of their opposite angles to the tangent of half their difference.
b2 -{- c2 — a2
Ey Art, 16, cos A = ^ 5
a2 — b2 = c2—2bccosA — C 02 — 2b cos A),
{a 4* b) (a — b) = C (c — 25 cos A) = c (AD ^ DB), c : a 4- b : a — b : AB — DB.
From this we have the following rule : Take the longest side for the base; then as the base is to
the sum of the other two sides, so is their difference to the difference of the segments of the base.
Half the sum of the segments added to half their difference gives the greater segment, and half their difference subtracted from half their sum gives the least.
The following logarithmic forms for right-angled triangles may also be useful, although they are only the definitions at page 8 put into logarithms.
Referring to the figure, page 52, considering C to be a right angle, we have by the definitions,
90 TRIGONOMETRY.
- BO a A AC b , A BC a. sin A = — rz - , cosA=z—, ttm A=: — =2-. AB c AB c AC b JL^
in logarithms, log a = log c + log sin A — 10 = log b + log tan A — 10, log b nr log c + log cos A — 10 = log a — log tan A — 10, log'c = log a — log sin A + 10 = 10+ log a — log sin A.
HEIGHTS AND DISTANCES, ETC. (1) In the year 1784, a hase BC being measured on
Blackheath, of a mile in length, the angles of elevation of Lunardi's balloon were taken, at the same time, by observers placed at its two extremities and in the middle; the one at c ^ B being 46° 10', that at .P550 8', and that at C 54o30#: required the height OA of the bal¬ loon.—Eonnycastle's Trigonometry.
Investigation. Ey right-angled triangles,
AB = OA cot OB A; AP— OA cot OP A ]AC=i OA cot 0 CA. IsTow AB2 + AC2 = 2AP2 + 2BP2,
or AB3 + AC2- 2AP2 = 2BP29
hence by substitution we have OA2 cot2 OB A + OA2 cot2 OCA —2 OA2 cot2 OPA = 2BP2
i
OA2 (cot2 OB A + cot2 OCA —2 cot2 OPA) = 2BP2}
2BP2
.\ OA2-
OA =
cot2 OB A + cot2 OCA —2 cot2 OPA9
BPV2 ■ Vfcot2 OBA + cot2 OCA — 2 cot2 OPA} '
(2) Observing an object OA at a distance, I took its angle of elevation OBA at the place where I stood, and found it to be 50° 23'; I then mea- ^:r sured a distance PB of 60 yards, in the most c ^.--'^>7 convenient direction the ground afforded, and ^ at this station found its elevation OPA to be 40° 33'; after which I measured on, in the b
It!
TEIG ON" OMETIiY. 97
same line, 50 yards further to C, and at this place fqund its elevation OCA to be 30° 49'; from which it is required to determine the height of the object, and its distance from each of the three stations 2?, P, C.
Investigation.
Since= OA cot OJBAiAP= OA cot OPA, A C= OA cot 0 OA
BP*+AP2—JL& z 20s BP A = 2AP BP ' an cos which is equal t<
CP* + AP* — AC* cos BP A 2AP.CP '
-BP2 + AP2 — AW _ CP*+AP* — AC2
"BP CP '
CP. AIP+BP. AC2-{CP+BP) AP2 = (OP+BP).CP. BP
by substitution we have
CP. OA2 cot2 OBA + BP. OA2 cot2 OCA—(CP + BP) x
OA2 cot2 OPA — (CP + BP) . CP. BP, this reduced gives
/CPcot2 OBA+BPcot2 OCA—( CP+BP) cot2 OPA OA-^/ (CP+BP). CP. BP
u (3) If the height of the mountain called the Peak of Teneriffe be 2! miles, as it is very nearly, and the angle taken at the top of it, as formed between a plumb-line and a line conceived to touch the earth in the horizon, or farthest visible point, be 88° 2'; it is required from these measures to determine the magnitude, of the whole earth, and the utmost distance that i can be seen on its surface from the top of the mountain, supposing the form of the earth to be perfectly globular ?
Let 1LF represent the height of the mountain, ED the tangent z. E = 88° 2', DC a semidiameter of the earth.
Since BBC is a right angle, and z. C = 1° 58', ^ CDF — ^ CFD, each of which = 89° 1', and ^ FFD = 90° 59', consequently FDF = 0o_59'.
P
98 TKIGOXOiTEXRY.
Then, as sin /_ EBF 0° 59' log ar. co. 9,765443 : sin i_ EFD 90° 59' log 9*999936
:: EF 2i miles log 0-367977 : FD 135-94 log 2-133356
And (Euc. III. 36) ^1'35 94^ 2-3333 = 7916 miles. 2-3333
(4) A point of land was observed, by a n ship at sea, to bear E. by S. ; and after sail- ing IST.E. 12 miles, it was found to bear S.E. wr by E. It is required to determine the place of that headland, and the ship's distance from s it at the last observation.
Let C be the point of land; A the former and B the latter place of observation. In the triangle ABC are given the angle A 56° 15', the angle B 101° 15', and the side AB 1~2 miles; consequently the angle C is 22° 30'.
Then, as sin 22° 30'"^ C log. ar. co. 10-417160 : sin 56° 15' Z. A log 9-919846
:: 12 miles AB log 1-079181
: 26-0728 miles BC, distance required, log 1-416187 _ (5) A string passes round the circumferences of two given
circles, in the same plane, and at a given distance from each other: required the length of the string.
Let GEIFBH represent the string touching the two circles, draw from the centres B and O the radii BB, OF, to the points of tangence; these then being each perpendicular
D »
E G to FB are parallel to one another; through F draw FC parallel to OB, then FC = OB and FO = B C ;
Let OB — 14 in., OF = 1 in.., and BB = 8^- in. Now FB — \/FCl — BC1 = \/OBl — (BB — OFf '
= V(142 — 7-53) = 11-82 1 59 : , . FB FB 11-82159
and sm FCB = — = — = = -844399 ; IC OB 14
TRIGO^OMETEY. 99
Z. FQD = OBQ = z IOF = 57° 36' 25". Hence z. DBH— 180° — z. BBO = 180° — 57° 36' 25"
= 122° 23' 35". Then, as 360° : 57° 36' 25" :: 6*2832 (circumference of
the less circle) : 1-00543 = the arc FL Also 360° : 122° 23' 35" :: 53*4072 (larger circum.) : 18*15742 = arc DH\
Hence IF + FD + BK — 30*98444, twice which, or 61*96888 inches = 5 ft. 2 in. nearly, is the length of the band.
(6) The side of a hill forms an inclined plaiie whose angle of inclination is known; required the direction in which a rail or other road must run along the side of the hill, in order that it may have an ascent of 1 in every n feet.
(7) ^Wanting to know the height of an inaccessible tower; at the least distance from it, on the same horizontal plane, I took its angle of elevation equal to 58° ; then, going 300 feet directly from it, found tKe angle there to be only 32°; re¬ quired its height, and my distance from it at the first station ?
Height 307*53, distance 192*15 ft* (8) Eeing on a horizontal plane, and wanting to know the
height of a tower placed on the top of an inaccessible hill: I took the angle of elevation of the top of the hill 40°, and of the top of the tower 51°; then measuring in a line directly from it to the distance of 200 feet farther, I found the angle at the top of the tower to be 33° 45'. ^WTiat is the height of the tower? 93*33148 feet.
(9) Erom a window near the bottom of a house, which seemed to be on a level with the bottom of a steeple, I took the angle of elevation of the top of the steeple equal 40°; then from another window, 18 feet directly above the former, the like angle was 37° 30': required the height and distance of the steeple? Height 210*44, distance 250*79 ft.
(10) "Wanting to know the height of, and my distance from, an object on the other side of a river, which appeared to be on a level with the place where I stood, close by the side of the river; and not having room to measure backward, in the same line, because of the immediate rise of the bank, I placed a mark where I stood, and measured in a direction from the object, up the ascending ground, to the distance of 264 feet, where it was evident that I was above the level of
p 2
100 TEIGONOMETSY.
the top of the object; there the angles of depression were found to be, viz., of the mark left at the rivers side 42°, of the bottom of the object 27°, and of its top 19°. Bequired the height of the object, and the distance of the mark from its bottom? Height 57*26 ft., distance 150*50 ft.
(11) Wanting to know my distance from an inaccessible object 0, on the other side of a river, and having no instru¬ ment for taking angles, but only a chain or cord for measuring distances, from each of two stations, A and B, which were taken at 500 yards asunder, I measured in a direct line from the object 0, 100 yards, viz., AG and BB each equal to 100 yards; also the diagonal AB measured 550 yards, and the diagonal BC, 560. "What was the distance of the object from each station A and B ? AO 536*81 yds., i? 0500*47 yds.
(12) The elevation of a tower is observed. At a station (a) feet nearer, the elevation is the complement of the former; (5) feet nearer still, it is double the first elevation. Shew that the height is
(13) If the angles A, B, C of a triangle ABO be as the .numbers 2, 3, 4 respectively; then
A a + c 2 cos — = ——.
2 b (14) If in the three edges which meet at one angle of a
^ube, three points A, B, O be taken at distances a, 5, c from the angle respectively; the area of the triangle AB O formed by joining the three points with each other
= \ ^(a^ + aV + JV).
(15) If the tangents of the angles of a plane triangle be in a geometric progression, whose ratio is n, shew that
sin 2 O = n sin 2A. (16) If the tangents of the semiangles of a plane triangle
be in arithmetical progression, the cosines of the whole angles will also be in arithmetical progression. If JR, and r be the radii of the circles described about and inscribed in the triangle ABO\ then the area of the triangle
A = Br (sin A + sin B + sin O).
TRIG ONOMETILY. 101
(17) In a right-angled triangle, GD^ CJE are drawn from the right angle (7, making angles a, (5 with the hypothenuse; the area of the triangle
(fir GBB= — (cota-cot/3).
JdC" (18) If a, y denote the lines bisecting the opposite
sides a, 3, c of any plane triangle, shew that
«< + /34 + -/ = 1 (^ + ^ + ,») ;
a2y33 + /32y2 + y2a2 = ^ (arb'2, + Z>V + cV).
(19) Let «, 5, 0 be the sides of a plane triangle, and a, /3, y the lines bisecting the angles and terminating in the oppo¬ site sides; then will
a/3y 4 [a + b + c) „, -V- = — x area of the triangle. ahc (ia + b) (b + c) (c + a)
(20) If lines be drawn from the angles of a plane triangle to a point within it, so as to make equal angles with each other, their sum will be represented by
-/{a3 - 2ab cos (C+ 60°) + b2}. (21) The squares of the sides of a plane triangle added to
the squares of the radii of the four circles respectively touch¬ ing these sides is equal to four times the square of the circum¬ scribing diameter; required a demonstration.
(22) If ^ b, c be the sides of a triangle, and a, /3, y the perpendiculars upon them from the opposite angles, shew that
a2 /32 y2 be ac (lb l Jl U—= H . Py ay aft d2 52 &
(23) If R be the radius of the circle circumscribed about a given triangle AB C, and r the radius of the inscribed circle, shew that
i ■ 5=i 11 - (1Bin^ + sin3!'+sinJ ? ))'•
(24) In 'any plane triangle, the square of the distance
102 TRIG ON02LETRY.
between the centre of the inscribed circie and the intersection of the perpendiculars
a3 = 4i22 — 2Rr — ——r———. a + o + c
(25) ABC is a plane triangle, P a point within it from which the sides subtend equal angles, it is required to find any number of such triangles having the sides, and the dis¬ tances AP, UP, CP all expressed in whole numbers. (See fig. page 71).
(26) If T (1) denote the sum of the angles of any polygon, T (3) the sum of all the products that can be made by taking them by threes, T (5) the sum of all the products that can be made by taking them by fives, and so on, then supposing radius to be unity
T(l)-T(3) + T (5) - T (7) + &c. = 0.
(27) The sum of the diameters of the inscribed and circum- -scribing circles of any plane triangle is equal to
a cot A + h cot JB + c cot C.
(28) If a, 5, c denote the sides and S the semiperimeter of any plane triangle, the square of the line- drawn from the vertex to the lower extremity of the vertical diameter of. the circumscribing circle is
be {b + tf)2
4S(S-a)'
(29) In a plane triangle let r be the radius of the inscribed circle R, It and E' the radius of the three circles touching the sides of the triangle externally; then will
I-L+I+I
required a demonstration.
(30) Let a, 5, c be the three sides of a triangle, and r r', r", r'" the radii of the four circles which touch them, then
al+ *0 + 10 = rr' + »t" + rr" + rY + rV + /V", that is, the sum of all the rectangles made bjr the binary combinations of the sides is equal to the sum of all the rect-
TEIGONOaiETItr. 103
angles made by the binary combinations of the four radii of the circles of contact.
(31) If from one of the angles of a regular polygon of n sides, lines be drawn to all the angles, shew that their
90° sum = a cosec2 —, 2a being a length of the side of the
n polygon.
(32) Having given three lines drawn from any point within a square to three of its angular points, determine a side of the square.
(33) The shadows of two vertical walls, which are at right angles to each other, and are a and aj feet in height, are observed when the sun is due south, to be I and bx feet in breadth; shew that if a be the sun's altitude, and /3 the inclination of the first wall to the meridian,
(34) The distance between the centres of two wheels = a, and the sum of their radii = c: shew that the length of a string which crosses between the two wheels and just wraps round them
(35) If an equilateral triangle have its angular points in three parallel straight lines, of which the middle one is dis¬ tant from the outside ones by a and 3, its side will be expressed by the following formula, viz.,
required the proof. (36) The hypothenuse (6) of a right-angled triangle
AJ3 C is trisected in the points D, JE: prove that if CD, CE bo joined, the sum of the squares of the sides of the triangle
CDS = — 3
104 TRIGONOMETRY.
(37) Pour places situated at unequal but given distances in the same straight line, appear to a spectator in the same plane with them to he at equal distances from each other; find the position of the spectator.
(38) The area of a quadrilateral inscribed in a circle =V{(£— a) (S—b) (S—c) (S — d)}.
(39) If D and D' be the diagonals of a quadrilateral figure, and 0 their angle of intersection, then the
area = ^ D.D' sin 0. Jj
CBAPTEB Y.
DEMOIVKE's TnEOREM.
33. This theorem, to which we give the name of the French geometer that discovered it, is as follows :
(cos 0 -f- -y/— 1 sin 0)n = cos ncp + ^/— 1 sin w0 (A). It expresses that, to raise the binomial
cos 0 + a/— 1 sin 0, to any power whatever, it is sufficient to multiply the arc 0 by the index of the power, we may either put the sign
+ or — before — 1, for that is the same as changing 0 into — 0.
The case where the index is a positive whole number is the first we shall proceed to consider.
Ey multiplication we find (cos 0 + y — 1 sin 0) (cos Y' + \/ — 1 sin Y')
= cos 0 cos — sin0sin Yr + \/— 1 (8^ 0008^ + cos0sin^). iNbw from the formula) Art. 13, the real part of this pro¬
duct is equal to cos (0 + ty), and the imaginary part is equal to y- 1 sin (0 + Y'); then.
(cos0 + V— 1 sin0) (cos Y^ + \/ — 1 sin YO = cos (0 + YO + V — 1 sin (0 + Y-).
TRIGONOMETRY. 105
That is to say, by multiplying together the two expressions ' of the form cos 0 + y/— 1 sin 0, we obtain a similar expres¬ sion, which contains the sum of the two arcs.
In order to multiply the product by a new factor of the same form, it is sufficient to add the new arc to the other two, and thus we may continue the operation to any number of factors. Then, if we suppose that there are n factors, each equal to cos 0 + V— 1 sin 0, it will become
(cos 0 + V— 1 sin 0)n= cos 720 + — 1 sin w0 (1).
Let us consider the case where the exponent is fractional.
In replacing 0 by the formula (1) becomes
0 . 0\* cos 1- y— 1 sm — j = cos 0 + ^ — 1 sin 0,
and, by extracting the nth root, and putting a fractional index in the place of the radical sign, the formula (A) will
be demonstrated for the index —; for we have n
(cos 0 +\/— 1 sin 0)re = cos - + \/~l sin - (2). n n K J
m Generally, the expression signifies that we ought to
raise A to the mth power, and afterwards take the nth. root of the result. Consequently, if we raise cos (p + sin 0 to the power m by the formula (1), and if afterwards we extract the nth. root by the formula (2), it will become
(cos 0 + V — 1 sin $) n = cos — + sin — (3). n n v i
This is the formula (A) in which n is changed into any
positive fraction — • n
Finally, when the index is negative, we observe that
(cos w0 -f- — 1 sin ?i0) (cos w0 — \/—l sin «0) = cos2 w0 + sin2 w0 = 1 ;
r 5
106 TRIGON 03IETEY.
and from this we find, 1 . . = cos M0 — s/ — 1 sin «0;
cos n<p + J— 1 sm nq> or which, is the same thing, (by formula 1),
(cos 0 + k/ — 1 sin 0) n =. cos (— ncp)
+ sin (— «0) (4), since the cosine of a positive arc is the same as the cosine of an equal arc taken negatively.
Thus, the formula (A) is true, when we take for n any positive or negative number whatever.
"We have not used irrational indices seeing that they are not of any utility, as we cannot replace them by commen¬ surable numbers, which however may differ therefrom as little as we please. And as to imaginary indices, they are not sus¬ ceptible of any interpretation.
The formula (.A), which is so simple and elegant, appears to have a serious defect when the index is fractional. In fact, the first member, being then equivalent to a radical expres¬ sion, ought to have many values, whilst the second member presents only one. The explanations which follow have for their object to correct this imperfection.
Let us return to formula (2) in which n is a positive whole number. From the principles of Algebra, the first member, which is equivalent to y(cos0-f- V— lsin0), ought then to have n different values; and in order that the second may give all of them, we shall shew that it will suffice to replace 0 by all the arcs which have the same sines and cosines as 0 itself. The general expression of these arcs is 0 + kC, C designating the whole circumference, and h any whole posi¬ tive or negative integer number. By putting 0 + IcC in the place of 0, the second member of formula (2) becomes,
0 +kC — . 0 + £0 ; cos -—5 Ys/ — 1 sm (5) • n n
And in this state, we say that it has precisely the same values as the first member.
And first, since n is an integer, it is clear, from formula (1), that by raising this second member to the power w, we arrive again at the form cos 0 + \/ — 1 sin 0.
TRIGONOMETRY. 107
In the second place, if we make successively 7c = 0, Ic = 1, 7; = 2 h = h — 1, we obtain n different values.
In fact, let there be any two values, 0 + aC — . 0 a0
cos */ •— 1 sm > w n 0+ 30 . $ + 30
and cos [- J — 1 sm 5 ft n
in which a and /3 are integer numbers Z. n- In order that they may be equal, it is necessary to have separately, an equality between the real and the imaginary parts; then the
difference of the two arcs ^ ail(^ $ P® ought to be n ft equal to one or more circumferences: but that difference,
which is^^—> is Jess ^a11 seeing that a and are ^ n :
consequently the two values are unequal. In the third place, if we take for h other numbers than
0, 1, 2 ft — 1, we shall find new values. In fact, all the other positive or negative integer numbers may be repre¬ sented by the formula fty + ft', y being any integer number, positive or negative, and ft' a positive integer Z. n; moreover by making k = fty ft' the expression (5) becomes
cos / „t</> + riC\t . / 0 + riC\ \yc+— )+ V—1 sin(y<7+ - j'
or rather, suppressing the useless circumferences, 0 + ft' (7 —- . / 0 + ft 0
cos — h V—1 sin
and as n is a positive number ^ ft, that value is comprehended under those we have obtained bv putting
7J=0, 1, 2 ft—1. Thus the second member of the formula (2) will acquire the generality, which it ought to have, if we take care to have for the arc 0 not only the arc 0 itself, but also the arcs
0 4" 0 4" 2 Cj 0 + (ft — 1) C. Tormula (3), ought also to be interpreted in an analogous manner, as wo find by raising cos 0 + V^l sin 0 to the
108 TRIGONOMETRY.
ttitli power, and extracting the nth root of the result. The formula (1), relating to the case of a whole positive index, gives
(cos 0 + >/— 1 sin 0)nl = cos ^— 1 sin M0, and afterwards, for the nth root, the formula (2) gives
(cos 0 + V— 1 sin 0)n = cos + V — 1 sin —• n n
But in order that the second member may have the same extension as the first, we mu^t, after what we have already- explained, put m<p + h C in the place of m(j); or, which is the same thing, replace mcfi successively by
m0, M0 + (7, m0 + (n — 1) C. 34. Formulae for sin m0, cos «0, and tan w0. Beferring
back to Demoivre's formulae,
cos w0 + ^ —1 sin w0 = (cos 0 + n/ — 1 sin 0)n (1), in which we suppose n to be a positive integer. Changing 0 into — 0 it becomes
cos nrf)— */ —I sin w0 = (cos 0 — \/ — 1 sin0)n (2). Adding and subtracting (1) and (2),
cos n<fi = (cos 0 + n/—1 sin 0)" + (cos <p - «y~l sin 0)" 2
sin n.<t> = (cos 0 + -/—I sin 0)"—(cos 0 — </—1 sin 0)" 2
Expanding by the binomial theorem, and suppressing the terms which destroy each other, we obtain
cos ^0 = (cos 0)n—2^(C0S <^)w~2 (s^n ^)2
+"(—"(';-2'3";-3)(co,0)-'(sm0).-te...(5),
,n i . i —l)(w — 2), fN . ,vo sm ncj) = n (cos0)n~1 sm 0 —12% ^(cos 0)n~3r(sin 0)3
n(n—\)(n — 2)(n—3)(n—4), . + 1.2,3.4.5 (cos ^ (sm 0)5—&c.. .(6).
TIUG ON OMETHY. 109
These formuloe express the sine and cosine of the multiple in terms of the sine and cosine of the simple arc 0: the
law of the terms is evident, and in the same manner as the binomial formuloe from which they are derived, the terms must be continued until we find a term equal to*nothing.
To find tan ncj), we must divide (6) by (5), , ^ n(n-iy(»-2)A ^ 0
sin n tan 0 , 2 q tan3 0 - &c. -z — tan n<p = cos iicp r
1 - \ 2 ; tan2 0 + &c.
35. Development of-the sine and cosine in a series. From equations (5) and (6), Euler has deduced the series
which expresses the sine and cosine of an arc in terms of the arc itself.
Betaining the supposition that n is a whole number, we can dispose of 0 in such a manner that ?z0 may be equal to
sequently the equations (5) and (6) will become x (x—0) „ ./sin0\2
cos a;=(cos 0)" 12 (cos 0)»-2 j
(1)_
sin»=»(»»«■-' ('iiSj_,.j,— (SSJtJ
x(x—0) (x—20) (a—30) (x-4<p) /sin 0\5 . + 1.2.3.4.5 (.cos?)j \~0~)
- &c....(2) Let us now conceive 0 to diminish to zero and n to increase
to infinity, then the above formulae will preserve no trace either of 0 or of w, and will contain the arc x only.
"When 0 becomes zero, we have . , sin 0
cos 0 = 1 and also - , = 1. 0
Admitting that the powers of cos 0 and of be also
equal to unity, however great the indices may be, the above formula) become
110 TRIG OK"03IETEY.
cos X— 1 lt2 + 1.2.3.4 1.2.3.4.5. 6"^ &c-"(3)>
sinii;=.i; — + i.2.3.4.5 - 1.2.3.4.5.6.7 &c-—(4)-
The following very sensible observations on these series are made by a correspondent of the Mechanics'' Magazine. (See No. 1256, pages 231, 232.)
"Now to many students, who are not altogether without thinking what they are about, and yet have not attained to clear notions on the distinction between things themselves and the numeftcal measures of those things—the above series are a sad stumbling-block, and well they may be. The reader who has these misty notions, naturally asks—4 How can a sine or cosine, which I know to be a number or ratio, be expressed in terms of an angle, which I conceive to be an opening ?1 How indeed! About the ne plus ultra of all con¬ ceivable absurdities would be the following equation ; sine of a certain opening
= that opening — g (cube of that opening)
"The absurdity arises from considering the symbol (0) as standing for the Euclidian angle or opening, instead of for a certain numerical fraction or ratio which measures that angle, and which bears the same proportion to other fractions (the measures of other angles), which the Euclidian angles themselves do. Cos 0 is really an abridgment for ' cosine of that Euclidian angle the length of whose arc bears the same proportion to the length of the radius, as the number 0 does to unity.'
'4 Again, the use of these series is seldom seen, and because the series on the left hand side of the equation is infinite in the number of its terms, the student is apt sometimes to con¬ ceive that an immense number of those terms must be taken to get anything like a tolerably near value for cos 0 or sin 0. To obviate this fancy, and at the same time to shew ckarly the real nature of the equation, let us take a numerical ex¬ ample and work it out. Suppose, then, we endeavour to find the numerical value of the cosine of 30 degrees by means of the series. Here (0) means
TBIGOX OlIKTRY. Ill
length of arc subtending the angle of 80° length of radius '
and this is the first thing to find. " Now, we know that the length of the circumference of a
circle bears the proportion to its diameter of (tt) to unity ; tt being nearly equal to S'HIG. "\Vc have then
^ D lire AB 30° 1 . / b ~~l — 9(P ~ 3 ' - TT
2
arc AB = ^ X a quarter of the circum- O
ference, =
arc AB radius ~
semi-circumference , . _ ,
1 semi-circumference 6"
3-1416 6 ' radius 6
so that here (0) is equal to •5236. " Take only the first three terms of the series for cos 0,
so that
cos 30° =1—7; (•5236)3 -f (•5236)4.
Keep only four places of decimals in multiplying, and the second side becomes
= 1 - 0-137078 + 0-003130 -= -866052. 4 'Now turn to a table of natural sines, &c., and you will
find the cosine of 30° to be -8660254, which agrees with that just found from the series up to the fifth place of decimals, and differs less from it than the cosines of 30° 1', or 29° 59', differ from the cosine of 30°."
36. In the applications of the higher branches of mathe¬ matics we often find it necessary to express (sin 0)" and (cos 0)n in terms of the sine and cosine of the simple arc. "We proceed in the following manner :
Let cos 0 + </— 1 sin 0 --= x,
then cos 0 y- 1 sin 0 = -•
n
112 JEIG OKOHETEY.
By addition, 2 cos 0 = x + -> x
and by subtraction, 2 s/— 1 sin 0 = as--;
x
.2n (cos 0)n = ^cc + i^
= + «^n~2 + n 2n~4 + ... -- \ 2 / x11'
= (collecting into pairs the terms equidistant from each ex¬ tremity of the series)
(^+j„)+«(^-2 + ii-3) +»(^) (*«-< +Jl) + &c.
but {cos 0 + V— 1 sin 0}n = cos ;*0 y/— 1 sin ncj) == ccn,
[ cos 0 — >/—1 sin0}n= cos«0 — v7— 1 sinw0 = ^>
by adding and subtracting, ^ye have
I 2 cos h0 = a?" + i;, and 2^/—1 sin «0 = a:'1 - i ;
1. (-" + ^) + » + » V (^-4 + J=r) + &c.
91 1 = 2 COS 720 + 2 W. COS (71 — 2) 0 + 2 w cos (72 — 4) 0, &c. ;
/. 2*"5 (COS 0)" = COS 720 + 72 COS (72 — 2) 0 72—1 .
+ 72 —COS (72 — 4) 0, &c.;
the last term of the series being 1 / 72—1\ 72 — ^72+1 2 n J ... p , when n is even,
, /72-l\ 72-i(72-l)+l and n y—J ... — ^ (?& — i) cos when 72 is odd.
Prom what has been given above, we have for the sine
2s/—1 sin0 = x — and 2^— 1 sin720 = xn— X X1
TRIGONOMETRY. 113
(2^-1 sin 0)- = (*-1)"= + (- l)y
zx*—nzn~- -{- n ^n~ ^ a:"-4... {a) 2 a:-
={*"+(it)} -" f" + {-rf}
wlien is even the series (b) becomes
(— 1)^ 2tt sin" 0 = ^cc" + — n (xn~2 +
, n — l/^i, 1\ + n —-— (as"-4 -| ) — ... 2 \ ^n-7
; 2 i»
= 2 cos n<j>—n. 2cos (?! — 2) 0 + » M ~ 1 2 cos — 4) 0 — 2 *
-f (— 1)1 nn~l n — jn + 1 _ 1 2 ^ n
( 1)J2"-1 sin"0 = cos «0 — w cos (n — 2) 0
+ n- —COS (ll — 4) 0 — ... Jj
+(-l)4»,i=i...!l=iZL±J. A JL /ii \n 71—1
If n be odd, then (v' - 1)" = J— 1 (J— 1= f — 1VTT. and the last term of (i) is ' K ' V '
(_!)«»-:) „»zJ » — *(»—0+1/ 1\ 2 i(» —1) K~~xr
114 TEIG0N03IETET.
Hence, when n is an odd number (&) becomes
( —1)~ T.\J — 1 sin" 0 — {xn — ^ — n (x*-2 -
•■••••< -- - '-|;:=:i+1 (—3-
=z 2 \/ — 1 sin720 — n.2\/ — 1 sin (m — 2)0 +
"jzl n — 1 n — bin—1) + 10 / r • ^ +(-1)2 !-(»_/) W-lsm0;
n—1 (— 1)3 2n~l sinn 0 = sin w0 — w sin (:n — 2) 0
+ n n~~21 s^n (w—4) 0 —
, / -.x^1 —1 n — hin — 1) + 1 • ,
(1) Sum the series
cos 0 + cos 2 0 + cos 3 0 + &c. + cos ??0.
Tha sum is equal to the two following series, . .
^ {x + ar + + &c. + a"),
2\x or or 1 x71)
. _ * , («■—in 2 \ cc— 1 cc71 (a:^— I)/
, j« i w4-1 1 n) (V + ^r)
l(K--l)(A'n+1 + l)__ 1 v rS/ \ aPi-/ — 2 x'lx— 1) 2 ,1 ^ —T- a:4
TRIG OXOMETHY. 115
T 1 ^ 1 -v- x-—— far rV
Now,
= /+J)'-4 /bih; V (1/I + J.)*_4 V (2C.»|)'-.
- 9 n& • 1 — cos2 — sin — _ / 2__
A/ - o ^0 ~~ . " v 1—cos-^f sin -- 2 2
consequently the sum of the series is equal to
. 9id> sin —
sin^^ 2
(n'+ X cos ( —
(2) Sum the series cosec 0 + cosec 2 0 cosec 23 0 to ft terms.
cosec 0 = cot — cot 0,
cosec 2 0 = cot 0 — cot 2 0, &c.
cosec 2n~10 = cot 2n~'2 0 — cot 2n""10;
hence, by addition,
cosec 0 + cosec 2 0 + &c. cosec 2n~10 = cot — cot 2,l~10.
(3) Sum tho series
1,01-01 (h 2tal12 + jp" 2r+ ¥ttin 2® + &C* t0 " tei'mS'
v Since tan^ = cot- — 2 cot ffi, 2 2
1IG TEIGONOMETHY.
1 (h 1 (h we nave - tan — = - cot — — cot 0,
2 2 2 2
1,0 1,0 1,0 , tan — = — cot — — - cot —,
22 22 22 22 2 2 &c
1 0 1 0 1,0 ~ tan — z=r — cot- -cot —r. 2n 2n 2n 2n 2n~1 2n~l
Bv addition ^ve have the sum
1 0 = — cot cot 0.
2* 2n
(4) Sum the series sin 0 + sin 20 + sin 30 + &c.
MISCELLANEOUS EXAMPLES.
(1) The square of the distance between the centres of two of1'the escribed circles of a triangle exceeds the square of the sum of their radii, by square of the opposite side of the triangle.
(2) Let a, b, c be the three distances of a point from three successive angles of a square field, then if S be the side of the square, and A the area of the triangle, whose sides are a, b js/2, c, it is required to prove that
S2 = l (a- + C-) ±2 A.
(3) The continued product of the radii of any three circles mutually touching each other, divided by the radius of the circle passing through their points of contact, gives the area of the triangle formed by joining their centres.
(4) Shew that if there be inscribed in a circle a regular figure, each of whose sides is an ?wth part of the radius, the secant of the angle at the centre subtending each side
2 m" 2)11*— 1 *
(5) Two regular polygons of the same number of sides being described, the one within, and the other without, the
TRIG ONOM-ETKT. 117
same circle, what will be the number of sides, when the whole space intercepted between the two polygonal boun¬ daries is an assigned part of either polygon? Ex. What is the figure when the exterior area is i of the interior polygon ?
(6) In a right-angled triangle, if the hypothenuse (e) be divided into segments (03), (y) by the line which bisects the right angle, and t = the tangent of half the difference of the acute angles,
x : y :: !+£: 1 — L
(7) To divide a given angle into two others, whose sines shall be in the ratio of m : n.
(8) A circle is inscribed in an equilateral triangle, an* equilateral triangle in the circle, a circle again in the latter triangle, and so^n; if r, r2) rs, &c., be the radii of the circles, shew that
r=:rl + ra + + &c.
(9) If by c be the sides of a triangle, and p, q, r perpen¬ diculars from a point within the triangle bisecting the sides, prove that
, {(t , b , c\ abc 4( —| 1— 1 — —. \p q r) pgr
(10) It is required to determine the continued product of n terms of the series
(sin 0 cos ^0)^ (sin^0cos^0)i (sin^ 0 cos
(Sia ^0003^0) 2*.
(11) The circumference of the inner of two concentric circles, of which It and r are the radii, is divided into n equal parts; shew that the sum of the squares of the lines drawn from the points of division to any point in the circumference of the outer circle = n (i22 + r2). '
(12) The ratio between the area of an equilateral and equiangular decagon described about a circle, and that of another within the same circle, is equal to
118 TRIG ONOlIETRY.
8
4 +
(13) In a plane triangle ABC, having given the sum of the sides A C, CB, the perpendicular from the vertex G upon the base AB; and the difference of the segments of the base made by the perpendicular; find the sides of the triangle.
(14) The length of a road in which the ascent is 1 foot in 5, from the foot of a hill to the top is a mile and two- thirds. What will be the length of a zigzag road, in which the ascent is 1 foot in 12 ?
(15) At each end of a horizontal base measured in a known direction from the place of an observer, the angle which the distance of the other end and a certain object subtends is ob¬ served, as also the angle of elevation of the object at one end of the base. Eind its height and bearing.
(16) A staff one foot long stands on the top of a tower 200 feet high. Shew that the angle which it subtends at a point in the horizontal plane 100 feet from the base of the tower is nearly O'-Sl".
(17) Pind the degrees, minutes, and seconds, in the angle whose circular measure is *1; also the values of cos mir and tan-1 (—1)TO (in an integer).
(18) Two circles have a common radius (r) and a circle is described touching this radius and the two circles: prove that
3r the radius of the circle which touches the three = —.
56 (19) At noon, a column in the E.S.E. cast upon the ground
a shadow, the extremity of which was in the direction jN^.E. : the angle of elevation of the column being a0
y and the distance of the extremity of the shadow from the column («) feet, determine the height of the column.
m Tm-, (-I)-*
TRIGONOMETEY. 119
(21) If A, i?, C be the angles of a triangle, ^ . nA . nil . nC
cos nA + cos nB + cos n C = 1 ± 4 sm — sin — sm —, 2* Jj 2* nA nB n C
or = — 1-1-4 cos — cos — cos —, — 2 2 2 according as n is odd or even.
(22) In a regular polygon of n sides inscribed in a circle whose radius is (r), if a be the distance of any point from the centre, and perpendiculars be drawn from this point upon all the sides of the polygon, the sum of the squares of the lines joining the feet of the adjacent perpendiculars is
7i sin2 ^ (r2 + a2), n
(23) Pour objects situated at unequal but given distances in the same straight line, appear to a spectator in the same plane with them to be at equal distances from each other, it is required to determine his position.
(24) If a', V be the segments of the hypothenuse made by a line bisecting the right angle, then
dV _ rt2-M2
ab {a + by (25) A boy flying a kite at noon, when the wind was
blowing a0 from the south, and the angular distance of the kite's shadow from the north was )30, the wind suddenly changed to ct^ from the south, and the shadow to from the north, and the kite was raised as much above 45° as it had before been below that elevation. Shew that 0° being the angular elevation of the sun, and 45° — 0° that of the kite at first,
^ o ™ sin P si11 P tan 0 ~ - — — sin (a —/3) sin (a, —/3,)' '
tan" (45° - 0°) = sm (a —P) sm v sm(a1—/Sj sm
INDEX.
Abbreviation of trigonometrical quj ntities. 8.
Angle, circular measure ot the, 3; divi¬ sions of the, 1. Angles, relation between the angles and sides of a triangle, 52; to find the sine and cosine of the sum and difference of two, 25.
Angular measure, described, 2; examples, 3; unit oi,2.
Arc. series for sine and cosine in terms of the, 109; remarks on the object and use of, 110; summation of the series, 114.
Area of polygons, to find the, 80. Area of a triangle, to find the, 56; am¬
biguous case, 58; to find the area of a triangle in terms of the radius of the inscribed circle, 59; to find the area of a triangle in terms of the radius of the circumscribing circle, 60; to find the radius of the circumscribing circle in terms of the sides, 60; when two sides and the included angle are given, 56.
Area of a triangle, examples, 60. Centisimal division, 1. Circle, ratio of circumference to diameter,
1; subdivisions 61 the, 2; examples, 2; to find the radius of a circle, described about a regular polygon of n sides, 80; to find the area of a polygon of n sides inscribed in a circle, 81; examples, 83.
Circumference of a circle, ratio of, to the diameter, 1.
Cosecant, definition of, 8; ratio of, 8{ value of, 300, 45®, and 600, 14, 15.
Cosine defined, 8. Cosine of 150, 180, 360, 540, 740,
values of, 14, 34,35; examples, 36; of 300, 450, 600, value of, 14; £ A, value of, 55; observations on, 55; formulae expressing the sin, cos. and tan. of the multiple n 0,108.
Cosine, ratio of, 8. Cosine and sine, series for, in the terms
of the arc, 109; remarks on the object and use of, 110; summation of the series, 114.
Cosine and sine of A B, to find the ex¬ pressions for, 26.
Cosine, to find the, and sine, of the sum and difference of two angles, 25.
Cosine 3 A, to find the expressions for, 28.
Cosine, to find, of an angle of a plane triangle in terms of the sides, 52; when the triangle is obtuse-angled, 52; when the triangle is acute-angled, 53.
Cotangent, defined, 8; ratio of, 8. Cotangent of 300, 450, 600, value of, 14,
15; (A ± B) to express in terms of cot. A and cot. B, 31.
Definitions, 8. Demoivre's theorem explained, 104. Diameter of a circle, ratio of, to the cir¬
cumference, 1. Distances, trigonometrical,calculation of:
formulse and examples, 96, English angular measure, described, 1;
method of reducing into French, 1. Euler's series for sine and cos. in terms
of the arc explained, 109; remarks on the object and use of, 110; summation, 114.
Examples, miscellaneous, 116. Formulae expressing the sine, cosine, and
tangent of the multiple n <f>, 108. Formulae of important equations to the
triangle, 55. Formulae of verification, 33. French angular measures described, 1;
method of reducing to English, 1. Functions, trigonometrical, described, 33;
tracing the values of, 12; results, 15; examples, 15.
Heights, trigonometrical, calculation of: formulae and examples, 96.
Inveree trigonometrical functions de¬ scribed, 32.
Miscellaneous examples, 116. Measure, angular, examples of, 3; English
angular, 1; French angular, 1; method of reducing one into the other, 1; unit of angular, 2.
I Measure, circular, of the angle, 3.
122 INDEX.
Polygon, 80; to find the radius of a circle described about a regular, 80.
Polygon, to find the area of a, inscribed iu a circle, 81; examples. 83.
Eatios, trigonometrical, 8: abbreviation of, 8.
Secant, definition of, 8; ratio of, 8. Secant of 300, 450, 600, values of, 14,15. Sexagesimal division, 1. Sine, defined, 8; sine and cos. of A B, to
find, 26; sine and cos., series for, in terms of the arc, 109; remarks on the objects and use of the series, 110; sum¬ mation of the, 114; formulae expressing the sin. cos. and tan. of the multiple n <£, 108; 3 A, to find the expressions for, 27: 4 A, &c., to find the expres¬ sions for, addition, multiplication, ratio of, 8.
Sine of an angle =» the cos. of its comple¬ ment, 10.
Sines of the angles of a plane triangle are proportional to the sides which sub¬ tend them: formulse, 52.
Sines, to find the, and cosines of the sum and difference of two angles, 25.
Sines of 300, 450, 600, values of, 13,14; of 150,18°, 360, 540, 740, values of, 34, 35; examples, 36.
Sines, A, sin. \ A, values of, 55; observa¬ tions on, 55.
Tables, trigonometrical construction of, described, 76 ; examples, 79.
Tangent, definition, 8; formulae express¬ ing the sin. cos. and tan. of the mul¬ tiple n <J>, 108; ratio of, 8; (A B) to express in terms of tan. A and tan. B, 30; i A, value of, 55; observations on, 56.
Tangent of 300, 450, 600, values of, 14, 15.
Triangle, area of a, 56; when two sides and their included angle are given, 56; ambiguous case, 58; to find the, in terms of the radius of the inscribed circle, 59; to find the area, in terms of the circumscribing circle, and the sides, 60; to find the radius of the cir¬ cumscribing circle in terms of the sides, 60; examples, 60.
Triangle, plane, parts of a, described, 51; the sines of the angles of a plane triangle are proportional to the sides which subtend them, 52; to find the «osine of an angle of a plane triangle, in terms of the sides:—(1) when the triangle is obtuse angled, 52; (2) when the angle is acute angled, 53; relations between the angle and sides of a triangle, 52.
Triangles, solution of: remarks on the solution of, 75; rules on the solution of, 95; solution of, 51; examples, 83.
Trigonometrical functions, tracing the values of, through the four quadrants, 12; results, 15; examples, 15.
Trigonometrical functions, inverse, de¬ scribed, 32.
Trigonometrical ratios, abbreviation of, 8 ; values of:—(1) sine of an angle is = the cosine of its complement, 10; (2) to show that sin. A = sin. (1800 — A), 10; (3)sine (— A) = — sin. A, and cos. (— A) = cos. A, 11.
Trigonometrical tables, construction of, 76; examples,79.^
Trigonometry described, 1. Unit of angular measure, 2. Verification, formulse of, 33.
THE END.
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