trigonometric equations - GSM Group of Education

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GSM CLASSES,Subhash bazar,Tonk 9667754117

1. DEFINITION

The equations involving trigonometric function of unknown angles are known as Trigonometric equations

e.g. cos = 0 , cos2 – 4cos =1 , sin2 + sin = 2 , cos2 – 4sin =1

A solution of a trigonometric equation is the value of the unknown angle that satisfies the equation.

e.g.,1 3 9 11

sin or , , , .........4 4 4 4 42

2. PERIODIC FUNCTION

A function f(x) is said to be periodic if there exists T > 0 such that f(x + T) = f(x) for all x in the domain of definitions

of f(x). If T is the smallest positive real numbers such that f(x + T) = f(x) , then it is called the period of f(x)

Since sin (2n + x ) = sinx , cos (2n + x) = cos x ; tan (n + x) = tan x for all n Z

Therefore sinx , cosx and tanx are perodic function the period of sinx and cos x is 2 and that of tanx is .

Function Period

sin (ax + b) , cos (ax +b), sec (ax + b) , cosec (ax +b ) 2/a

tan (ax + b), cot (ax +b) /a

| sin (ax + b), | cos (ax +b) | , | sec (ax +b) | , | cosec (ax +b ) | /a

| tan (ax + b ) | , | cot (ax +b ) | /2a

3. TRIGONOMETRICAL EQUATIONS WITH THEIR GENERAL SOLUTION

Trigonometrical equation General solution

If sin = 0 then = n

If cos = 0 then = (n + / 2) = (2n+1)/2

If tan = 0 then = n

If sin = 1 then = 2n + /2 = (4n+1)/2

If cos = 1 then = 2n

If sin = sin then = n + (–1)n where [–/2 , /2]

If cos = cos then = 2n where ( 0, ]

If tan = tan then = n + where (–/2 , /2]

If sin2 = sin2 then = n

If cos2 = cos2 then = n

If tan2 = tan2 then = n

If sin sin

*cos cos

then = 2 n +

If sin sin

*tan tan

then = 2n +

If tan tan

*cos cos

then = 2n +

* Every where in this chapter "n" is taken as an integer.

TRIGONOMETRIC EQUATIONS

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* If be the least positive value of which statisfy two given trigonometrical equations , then the general value of

will be 2n +

4. GENERAL SOLUTION OF STANDARD TRIGONOMETRICAL EQUATIONS

Since, trigonometric functions are periodic, The solution consisting of all possible solutions of a trigonometric

equation is called its general solution.

We use of following results for solving the trigonometric equations ;

Result 1 : sin = 0 = n , n I .

We know that sin = 0 for all integral multiples of . (by graphical approach)

sin = 0 = 0 , , 2 , 3 , .......

= n , n I .

sin = 0

= n , n I .

Result 2 : cos = 0 = (2n +1) 2

, n I .

We know that cos = 0 for all odd multiples of 2

(by graphical approch)

cos = 0 = 2

3

2

,

5

2

, .......

= (2n +1)2

, n I .

cos = 0

= (2n+1)2

, n I .

Result 3 : tan = 0 = n , n I .

We know that tan = 0 for all integral multiple of .

tan = 0 = 0 , , 2 , 3 , .......

= n , n I .

tan = 0

= n , n I .

Result 4 : sin = sin = n + (–1)n , where n I and ,2 2

.

We have ,

sin = sin , where ,2 2

sin – sin = 0

2 cos 2

sin 2

= 0

cos 2

= 0 or sin 2

= 0

sin

2

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2

= (2m +1) 2

, m I or

2

= m , m I .

( + ) = (2m + 1) , m I or ( – ) = 2m , m I

= (2m + 1) – , m I or = (2m ) + , m I

= ( any odd multiple of ) –

or = (any even multiple of ) +

=n + (–1)n , where n I

sin = sin

= n + (–1)n , where n I and ,2 2

Result 5 : cos = cos = 2n , n I and [ 0 , ]

We have,

cos = cos , where 0 ,

cos – cos = 0

–2 sin 2

. sin 2

= 0

sin 2

= 0 or sin 2

= 0

2

= n or

2

= n , n I .

+ = 2n or – = 2n , n I

= 2n – or = 2n + , n I

= 2n , n I

cos = cos

= 2n , n I , where [ 0 , ]

Result 6 : tan = tan = n + , n I where ,2 2

we have

tan = tan , where ,2 2

sin sin

cos cos

sin cos – cos q sin = 0

sin ( – ) = 0

– = n , n I

= n + , n I

tan = tan

= n + where ,2 2

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Ex.1 Find the set of values of x for which 1x2tan.x3tan1

x2tanx3tan

Sol. We have, 1x2tan.x3tan1

x2tanx3tan

tan (3x – 2x) = 1 tan x = 1

tan x = tan 4

x = n + 4

{Using tan = tan = n + }

But for this value of x,

tan 2x = tan

2n2 , which does not satisfy the given equation as it reduces to indeterminate form.

Hence the solution set for x is .

5. GENERAL SOLUTION OF THE TRIGONOMETRICAL EQUATION

Result 7 : sin2 = sin2 , cos2 = cos2 , tan2 = tan2 = n

(i) sin2 = sin2

1 cos2 1 cos2

2 2

cos 2 = cos 2

2 = 2n 2 , n I

= n , n I

(ii) cos2 = cos2

1 cos2 1 cos2

2 2

cos 2 = cos 2

2 = 2n 2 , n I

= n , n I

(iii) tan2 = tan2

2 2

2 2

1 tan 1 tan

1 tan 1 tan

(applying componend and dividendo)

cos 2 = cos 2

2 = 2n 2 , n I

= n , n I

Ex.2 If x 2

n and 1)x(cos 2xsin3xsin2

, then find the general solutions of x.

Sol. As 2

nx

cos x 0, 1, – 1

So, 1)x(cos 2xsin3xsin2

sin2 x – 3 sin x + 2 = 0

(sin x – 2) (sin x – 1) = 0

sin x = 1, 2

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where sin x = 2 is not possible and sin x = 1 does not satisfy the equation.

no general solution is possible.

Ex.3 Solve 3 cos2 – 2 3 sin cos – 3 sin2 = 0

Sol. The given equation can be written as 3 tan2 – 2 3 tan– 3 = 0

2 3 12 36 1

tan , 36 3

Either tan = 1

3= tan

6

= n + /6 ....(i)

or tan = – 3 = tan ( – /3)

= n – /3 ....(ii)

n6

and n

3

Ex.4 solve sin 3x + cos 2x = – 2

Sol. Since, sin 3x –1 and cos 2x –1. We have, sin 3x + cos 2x –2.

Thus the equality holds true if and only if;

sin 3x = –1 and cos 2x = –1

3x = n + (–1)n

2 and 2x = 2n ± .

i.e., x =

6)1(

3

n n and x = n ±

2

, n

Solution set is,

2nx/x

6)1(

3

nx/x n

Note : here unlike all other problem the solution set consists of the intersection of two solution sets and not the

union of the solution sets.

6. GENERAL SOLUTION OF TRIGONOMETRICAL EQUATION a cos + b sin = c

To solve the equation a cos + b sin = c, put a = r cos , b = r sin such that

2 2 1 br a b , tan

a

Substituting these values in the equation we have r cos cos r sin sin = c

c

cosr

2 2

ccos

a b

If | c | > 2 2a b , then the equation ;

a cos + b sin = c has no solution

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If | c | 2 2a b , then put ;

2 2

| c |

a b = cos , so that

cos ( – ) = cos

2n

2n

Ex.5 Solve sinx + 3 cosx = 2

Sol. Given, 3 cosx + sinx = 2 , dividing both sides by 2 2a b

3

2cosx +

1

2sinx =

2 1cos x cos x 2n

2 6 4 6 42

x = 2n 5

x 2n , 2n4 6 12 12

where n I

Ex.6 The number of solutions of the equation sin 5x cos 3x = sin 6x cos 2x in the interval [0, ] are-

Sol. The given equation can be written as

)x4sinx8(sin2

1)x2sinx8(sin

2

1

sin 2x – sin 4x = 0 –2 sin x cos 3x = 0

sin x = 0 or cos 3x = 0. That is, x = n (n ), or 3x = k + /2 (k ).

Therefore, since x [0, ], the given equation is satisfied if x = 0, , /6, /2 or 5/6.

7. SOLUTIONS IN THE CASE OF TWO EQUATIONS ARE GIVEN

Two equations are given and we have to find the values of variable which may satisfy both the given equations,

like

cos = cos and sin = sin

so the common solution is = 2n + , n I

Similarly, sin = sin and tan = tan

so the common solution is , = 2 n + , n I

Rule : Find the common values of between 0 and 2 and then add 2n to this common value

Ex.7 Solve the following system of equation sinx + cosy =1 , cos2x – cos2y = 1

Sol. Given sinx + cosy = 1 ......(i) and (1 – 2sin2x) – (2cos2y –1) = 1 sin2x + cos2y = 1/2 .....(ii)

Put sin x = u and cosy = v in (i) and (ii) u + v = 1 u2 + v2 = 1/ 2

solving above equations u = 1

2 and v =

1

2 sin x =

1

2 x = n + (–1)n

6

, n I

cosy = 1

2 y = 2m

3

, m I the given equation have solution

x = n + (–1)n 6

, n I and y = 2m

3

, m I

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1. INTRODUCTION

The word 'Trigonometry' is derived from two Greek words

(1) Trigonon and

(2) Metron

The word trigonon means a triangle and the word metron means a measurement. Hence trigonometry means the

science of measuring triangles.

2. ANGLE

Consider a ray OA

. If this ray rotates about its end point O and takes the position OB , then the angle AOB has

been generated.

Vertex O

= angle

Initial side A

B

Term

inal s

ide

An angle is considered as the figure obtained by rotating a given ray about its end - point.

The initial position OA is called the initial side and the final position OB is called terminal side of the angle. The end

point O about which the ray rotates is called the vertex of the angle.

3. SENSE OF AN ANGLE

The sense of an angle is said to be positive or negative according as the initial side rotates in anticlockwise or

clockwise direction to get to the terminal side.

O

= +ve

Anticlockwise direction

A

B O = ve

Clockwise directionA

B

4. RIGHT ANGLE

When two lines intersect at a point in such a way that two adjacent angles made by them are equal, then each

angle is called a right angle.

90°90°

OX' X

A

5. A CONSTANT NUMBER

The ratio of the circumference to the diameter of a circle is always equal to a constant and this constant is denoted

by the Greek letter

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i.e. Circumference of a circle

Diameter of the circle (constant)

The constant is an irrational number and its approximate value is taken as 22

7. The more accurate value to six

decimals places is taken as 355

113.

6. SYSTEMS OF MEASUREMENT OF AN ANGLES

There are three systems for measuring angles

6.1 Sexagesimal or English system

6.2 Centesimal or French system

6.3 Circular system

6.1 Sexagesimal system : The principal unit in this system is degree (°). One right angle is divided into 90 equal

part and each part is called one degree (1°) . One degree is divided into 60 equal parts and each part is called

one minute. Minute is denoted by (1'). One minute is equally divided into 60 equal parts and each part is called

one second (1").

In Mathematical form :

One right angle = 90° (Read as 90 degrees )

1° = 60' (Read as 60 minutes )

1' = 60" (Read as 60 seconds )

Ex.1 40° 30' is equal to

(1)

o41

2

(2) 81° (3)

o81

2

(4) None of these

Sol. We know that , 30' =

o1

2

; 40° +

o1

2

=

o81

2

6.2 Centesimal system : The principal unit in system is grade and is denoted by (g). One right angle is divided

into 100 equal parts, called grades, and each grade is subdivided into 100 minutes, and each minute into 100

seconds.

In Mathematical form :

One right angles = 100g (Read as 100 grades)

1g = 100' (Read as 100 seconds)

1' = 100" (Read as 100 seconds)

Ex.2 25' is equal to -

Sol. 100' is equal to 1g

so is equal to

g g1 1

25100 4

Relation between Sexagesimal and Centesimal systems :

One right angle = 90° (degree system) ..... (1)

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One right angle = 100g (grade system) ..... (2)

by (1) and (2)

90° = 100g

or , D G

90 100

then we can say, ; 1° =

g10

9

, 1g =

o9

10

Ex.3 80g is equal to

Sol. We know that 1g =

o9

10

then, 80g =

o9

8010

80g = 72°

6.3 Circular system : In circular system the unit of measurement is radian. One radian, written as 1C, is the measure

of an angle subtended at the centre of a circle by an arc of length equal to the radius of the circle.

Consider a circle of radius r having centre of O. Let A be a point on the circle. Now cut off an arc AB whose

length is equal to the radius r of the circle.

B O r

Ic

r

r

C

A

In the adjacent figure OA = OC = arc AC = r = radius of circle, then measurement of AOC is one radian

and denoted by 1c. Thus AOC = 1c .

6.3.1 Some Important conversion

Radian = 180° One radian =

o180

6

Radian = 30°

4

Radian = 45°

3

Radian = 60°

2

Radian = 90°

2

3

Radian = 120°

3

4

Radian = 135°

5

6

Radian = 150°

7

6

Radian = 210°

5

4

Radian = 225°

5

3

Radian = 300°

6.3.2 Relation between systems of measurement of angles

D G 2C

90 100

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7. TRIGONOMETRICAL RATIOS OR FUNCTIONS

Let a line OA makes angle with a fixed line OX and AM is perpendicular from A on OX. Then in right-angled

triangle AMO, trigonometrical ratios (functions) with respect to are defined as follows :

sin = perpendicular(P)

hypotenuse(H)

cos = base(B)

hypotenuse(H)

tan = perpendicular (P)

Base (B)

cosec = H

P. sec =

H

B , cot =

B

P

Note :

(i) Since t-ratios are ratio between two sides of a right angled triangle with respect to an angle, so they are

real numbers.

(ii) may be acute angle or obtuse angle or right angle.

8. RELATIONS BETWEEN TRIGONOMETRICAL RATIOS

(i) 1 1 1

cosec , sec ,cotsin cos tan

(ii) sin

tancos

(iii) cos

cotsin

(iv) sin

2 + cos

2 = 1

(v) 1 + tan2 = sec

2 (vi) 1 + cot

2 = cosec

2

9. SIGN OF TRIGONOMETRIC RATIOS

(i) All ratios sinq, cosq, tan cotq, secq and cosecq are positive

in Ist quadrant.

(ii) sinq( or cosecq) positive in IInd quadrant, rest are negative.

(iii) tanq( or cotq) positive in IIIrd quadrant, rest are negative.

(iv) cosq( or secq) positive in IVth quadrant, rest are negative.

10. DOMAIN AND RANGE OF A TRIGONOMETRICAL FUNCTION

If f : X Y is a function, defined on the set X, then the domain of the function f, written as Domain is the

set of all independent variables x, for which the image f(x) is well defined element of Y, called the co-domain

of f.

Range of f : X Y is the set of all images f(x) which belongs to Y , i.e.

Range f = {f(x) Y:x X} Y

The domain and range of trigonometrical functions are tabulated as follows

O B

M

H P

Y

X

A

IInd Quadrant Ist Quadrant

IIIrd Quadrant IVth Quadrant

x-axis

y-axis

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Trigo. function Domain Range

sin x R, the set of all the real number – 1 sin x 1

cosx R –1 cos x 1

tan x R – 2n 1 ,n2

I R

cosecx R – n ,n I R – { x : – 1 < x < 1 }

sec x R – 2n 1 ,n2

I R – { x : –1 < x < 1 }

cot x R – n ,n I R

11. VARIATION OF VALUES OF TRIGONOMETRICAL RATIOS IN DIFFERENT QUADRANTS-

12. RELATION BETWEEN TRIGONOMETRICAL RATIOS AND IDENTITIES-

(1)

cos

sintan (2)

sin

coscot

(3) sin A cosec A = tan A cot A = cos A sec A = 1

(4) sin2 + cos2 = 1 or sin2 = 1 – cos2 or cos2 = 1 – sin2

(5) 1 + tan2 = sec2 or sec2 – tan2 = 1 or sec2 – 1 = tan2.

(6) 1 + cot2 = cosec2 or cosec2 – cot2 = 1 or cosec2 – 1 = cot2

Sine decreases from 1 to 0

cosine decreases from 0 to –1

tangent increases from – to 0

cotangent decreases from 0 to –

secant increases from – to –1

cosecant increases from 1 to

Sine increases from 0 to 1

cosine decreases from 1 to 0

tangent increases from 0 to

cotangent decreases from to 0

secant increases from 1 to

cosecant decreases from to 1

'

Sine decreases from 0 to –1

cosine increases from –1 to 0

tangent increases from 0 to

cotangent decreases from to 0

secant decreases from –1 to –

cosecant increases from – to –1

IV

Since increases from –1 to 0

cosine increases from 0 to 1

tangent increases from – to 0

cotangent decreases from 0 to –

secant decreases from to 1

cosecant decreases from –1 to –

'

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(7) Since sin2A + cos2A = 1, hence each of sin A and cos A is numerically less than or equal to unity i.e.,

|sin A| 1 and |cos A| 1

or –1 sin A 1 and –1 cos A 1

Note : The modulus of real number x is defined as |x| = x if x 0 and |x| = –x if x < 0.

(8) Since sec A and cosec A are respectively reciprocals of cos A and sin A, therefore the values of sec A

and cosec A are always numerically greater than or equal to unity i.e.,

sec A 1 or sec A –1

and cosec A 1 or cosec A –1

In other words, we never have

–1 < cosec A < 1 and –1 < sec A < 1.

While tanA and cotA may take any real value

13. TRIGONOMETRICAL RATIOS IN TERMS OF EACH OF THE OTHER

sin cos tan cot sec cosec

sin sin 2cos1

2tan1

tan

2cot1

1

sec

1sec2

eccos

1

cos 2sin1 cos 2tan1

1

2cot1

cot

sec

1

eccos

1eccos 2

tan

2sin1

sin

cos

cos1 2

tan cot

11sec2 1eccos

1

2

cot

21 sin

sin

2

cos

1 cos

1

tancot 2

1

sec 1 2cosec 1

sec 2

1

1 sin

1

cos21 tan

21 cot

cot

sec 2

cosec

cosec 1

cosec 1

sin 2

1

1 cos

21 tan

tan

21 cot 2

sec

sec 1

cosec

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14. TRIGONOMETRICAL RATIOS OF STANDARD ANGLES

15. TRIGONOMETRICAL RATIOS OF ALLIED ANGLES

16. GRAPH OF DIFFERENT TRIGONOMETRICAL RATIOS

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17. SUM AND DIFFERENCE FORMULAE

(i) sin(A + B) = sinA cosB + cosA sin B (ii) sin(A – B) = sinA cosB – cosA sin B

(iii) cos(A + B) = cosA cosB – sinA sinB (iv) cos(A – B) = cosA cosB + sinA sinB

(v) · tan(A + B) =BtanAtan1

BtanAtan

· tan(A – B) =

BtanAtan1

BtanAtan

(vi) · tan

tan1

tan1

4· tan

tan1

tan1

4

(vii) cot(A + B) =BcotAcot

1BcotAcot

(viii) cot(A – B) =

AcotBcot

1BcotAcot

(ix) sin(A + B) sin(A – B) = sin2 A – sin2B = cos2B – cos2A

(x) cos(A + B) cos(A – B) = cos2A – sin2B = cos2B – sin2A

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(xi) · sin2q = 2sinq cosq = )tan1(

tan22

· (cosA ± sin A)2 = 1 ± sin 2A

(xii) cos2q = cos2q – sin2q = )tan1(

)tan1(2

2

= 1 – 2 sin2q = 2 cos2q – 1

(xiii) · tan2q =

2tan1

tan2·

sin

)cos1( = tan

2

·

sin

)cos1( = cot

2

·

)cos1(

)cos1(

= tan2

2

·

)cos1(

)cos1(

= cot2

2

(xiv)A 1 cosA

sin2 2

, cos

A 1 cosA

2 2

(xv) tan A 1 cos A

2 1 coA

(xvi) sin 3A = 3 sin A – 4 sin3A or sin3 A = 1

4(3 sinA – sin 3A)

(xvii) cos 3A = 4cos3A – 3 cosA or cos3A = 1

4 ( cos 3A + 3cosA )

(xix) tan 3A = 3

2

3 tanA tan A

1 3tan A

( A n + /6 )

18. FORMULAE FOR TRANSFORMATION OF SUM OR DIFFERENCE INTO PRODUCT

(i) sinC + sinD = 2sin

2

)DC(cos

2

)DC(

(ii) sinC – sinD = 2cos

2

)DC(sin

2

)DC(

(iii) cosC + cosD = 2cos

2

)DC(cos

2

)DC(

(iv) cosC – cosD = 2sin

2

)CD(sin

2

)DC(

(v) tanA ± tanB = BcosAcos

BsinAcosBcosAsin

Bcos

Bsin

Acos

Asin

BcosAcos

)BAsin(

mB,

2nA

(vi) cotA ± cotB = BsinAsin

)ABsin(

2mB,nA

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(vii) · cosA ± sinA = 2 sin

A

4= 2 cos

A

4· tanA + cotA =

)AcosA(sin

1

(viii) · 1 + tanA tanB =BcosAcos

)BAcos( · 1 – tanA tanB =

BcosAcos

)BAcos(

(ix) · cotA – tanA = 2cot2A · tanA + cotA = 2cosec2A

(x) · sin2

A + cos

2

A = ± Asin1 · sin

2

A – cos

2

A = ± Asin1

19. FORMULAE FOR TRANSFORMATION OF PRODUCT INTO SUM OR DIFFERENCE

(i) 2sinA cosB = sin(A + B) + sin(A – B)

(ii) 2cosA sinB = sin(A + B) – sin(A – B)

(iii) 2cosA cosB = cos(A + B) + cos(A – B)

(iv) 2sinA sinB = cos(A – B) – cos(A + B)

20. TRIGONOMETRICAL RATIOS FOR SOME IMPORTANT ANGLES

(i) sin 1

72

=

4 2 6

2 2

(ii) cos

17

2

=

4 2 6

2 2

(iii) tan 1

72

= 3 2 2 1 (iv) sin15º =

22

)13( = cos75º

(v) cos15º = 22

)13( = sin75º (vi) tan15º = 2 – 3 = cot75º

(vii) cot15º = 2 + 3 = tan75º (viii) sin22º

2

1 = 22

2

1

(ix) cos22º

2

1 = 22

2

1 (x) tan22

º

2

1= 2 – 1

(xi) cot22º

2

1= 2 + 1 (xii) sin18º =

4

1( 5 – 1) = cos72º

(xiii) cos18º =4

15210 = sin72º (xiv) sin36º =

4

12210 = cos54º

(xv) cos36º =4

1( 5 + 1) = sin54º

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21. FORMULAE FOR SUM OF THREE ANGLES

(i) sin (A + B + C) = sinA cos B cosC + cosA sin B cos C + cos A cos B sin C – sin A sin B sin C

= cos A cos B cos C ( tanA + tan B + tanC – tan A tan B tan C )

(ii) cos (A + B + C) = cosA cosB cosC – sinA sinB cosC – sinA cos B sin C – cos A sin B sin C

= cos A cos B cos C (1 – tan A tan B – tan B tan C – tan C tanA )

(iii) tan (A + B + C) = tanA tanB tanC tanA tanB tanC

1 tanA tanB tanB tanC tanCtan A

(iv) · 4sin(60º – A) sinA sin(60º + A) = sin3A

· 4cos(60º – A) cosA cos(60º + A) = cos3A

· tan(60º – A) tanA tan(60º + A) = tan3A

22. CONDITIONAL IDENTITIES

(1) If A + B + C = 180° , then

(i) sin 2A + sin 2B + sin2C = 4 sin A sin B sin C

(ii) sin 2A + sin 2B – sin 2C = 4 cosA cos B sin C

(iii) sin (B + C –A) + sin (C + A – B) + sin (A + B –C) = 4 sin A sin B sin C

(iv) cos 2A + cos 2B + cos 2C = –1–4 cos A cos B cos C

(v) cos 2A + cos 2 B – cos 2C = 1 – 4 sinA sin B cos C

(2) If A + B + C = 180°, then

(i) sin A + sin B + sin C = 4cos A

2 cos

B

2cos

C

2

(ii) sin A + sin B – sin C = 4 sinA

2sin

B

2cos

C

2

(iii) cosA + cos B + cosC = 1 + 4 sinA

2sin

B

2sin

C

2

(iv) cosA + cosB – cos C = –1 + 4 cos A

2cos

B

2sin

C

2

(v)cos A cosB cosC

2sinBsinC sinCsinA sinA sinB

(3) If A + B + C = , then

(i) sin2A + sin2B – sin2C = 2 sin A sin B cos C

(ii) cos2A + cos2B + cos2C = 1–2 cos A cos B cos C

(iii) sin2A + sin2B + sin2C = 2 + 2 cosA cos B cosC

(iv) cos2A + cos2B – cos2C = 1–2 sin A sin B cos C

(4) If A + B + C = , then

(i) sin2 A

2 + sin2

B

2+ sin2

C

2=1 – 2sin

A

2sin

B

2sin

C

2

(ii)2 2 2A B C A B C

cos cos cos 2 2sin sin sin2 2 2 2 2 2

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(iii)2 2 2A B C A B C

sin sin sin 1 2cos cos sin2 2 2 2 2 2

(iv)2 2 2A B C A B C

cos cos cos 2cos cos sin2 2 2 2 2 2

(5) If x + y + z = 2

, then

(i) sin2 x + sin2y + sin2z = 1–2 sin x sin y sin z

(ii) cos2x + cos2y + cos2z = 2 + 2 sin x sin y sin z

(iii) sin2x + sin2y + sin 2z = 4 cos x cosy cos z

(6) If A + B + C = , then

(i) tanA + tan B + tan c = tan A tan B tan C

(ii) cotB cot C + cot C cot A + cot A cot B = 1

(iii)B C C A A B

tan tan tan tan tan tan 12 2 2 2 2 2

(iv)A B C A B C

cot cot cot cot cot cot2 2 2 2 2 2

(7) (a) For any angles A , B, C we have

(i) sin (A + B + C) = sin A cos B cos C + cos A sin B cos C

+ cos A cos B sin C – sin A sin B sin C

(ii) cos (A+B+C) = cos A cos B cosC – cos A sin B sin C

– sin A cos B sin C – sin A sin B cosC

(iii) tan (A + B + C) = tanA tanB tanC tanA tanB tanC

1 tanA tanB tanB tanC tanCtan A

(b) If A , B, C are the angles of a triangle, then

sin(A + B + C) = sin = 0 and

cos (A + B + C) = cos = –1

then (a) (i) gives

sinA sin B sin C = sin A cos B cos C + cosA sin B cosC + cos A cos B sin C

and (a) (ii) gives

1 + cos A cos B cos C = cos A sin B sin C + sin A cos B sin C + sin A sin B cos C

23. GENERAL SOLUTIONS OF TRIGONOMETRICAL EQUATIONS

*(i) If sin = sin then = n + (–1)n , n Z

*(ii) If cos = cos then = 2n , n Z

*(iii) If tan = tan then = n + , n Z

(iv) If sin2 = sin2

(v) If cos2 = tan2 then = n , n Z

(vi) If tan2 = tan2

(vii) If sin sin

cos cos

then = 2n + , n Z

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24. MOTHOD OF COMPONENDO AND DIVIDENDO

If p a

q b , then by componendo an dividendo we can write

p q a b q p b aor

q q a b q p b a

or p q a b q p b a

orp q a b q p b a

Note :- Reference of the above formulae will be given in the solutions of problems.

25. SOME IMPORTANT RESULTS

(i) 2 2 2 2a b asinx bcos x a b

(ii) sin2x + cosec2 x 2

(iii) cos2x + sec2 x 2

(iv) tan2x + cot2 x 2

(v)1 sin

tan sec tan1 sin 4 2

(vi)1 sin

tan sec tan1 sin 4 2

(vii)1 cos

cot cosec cot1 cos 2

(viii)1 cos

tan cosec cot1 cos 2

(ix) cos . cos 2 . cos 22 ............ cos 2n–1 = n

n

sin2

2 sin

; n

(x) cosA + cos (A +B) + cos (A + 2B) + ........ + cos { A + ( n –1) B } = sin nB / 2 B

cos A (n 1)sinB / 2 2

26. THE GREATEST AND LEAST VALUES OF THE EXPRESSION [ a sin + b cos ]

Let a = r cos ......... (1)

and b = r sin ......... (2)

squaring and adding (1) and (2)

then a2 + b2 = r2

or , r = 2 2a b

a sin + b cos = r(sin cos + cos sin )

= r sin ( + )

But – 1 sin 1

so –1 sin ( + ) 1

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then – r r sin ( + ) r

hence,

2 2 2 2a b asin bcos a b

then the greatest and least values of

a sin + b cos are respectively 2 2 2 2a b and a b

Least value of a sinx + b cos x + c is 2 2c a b and greatest value is 2 2c a b

27. MISCELLANEOUS POINTS

(i) Some useful identities :

(a) tan (A + B + C) = tanA tanA tanB tanC

1 tanA tanB

(b) tan = cot – 2 cot 2

(c) tan3 = tan . tan ( 60° –) . tan ( 60° + ) (d) tan (A+B) – tanA – tanB = tanA. tanB.tan(A+B)

(e) sin sin ( 60° – ) sin (60° + ) = 1

sin34

(f) cos cos ( 60° – ) cos (60° + ) = 1

cos34

(ii) Some useful series :

(a) sin + sin ( + ) + sin ( + 2) ......... + to n terms =

n 1 nsin sin

2 2; 2n

sin2

(b) cos + cos ( + ) + cos ( + 2) + ........ + to n terms =

n 1 ncos sin

2 2; 2n

sin2

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trigonometric equations - GSM Group of Education

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