Trigonometric Functions, Law of Sines and Law

of Cosines

Pedro J. A. Matos

Contents

1 Introduction 1

2 Trigonometric functions 22.1 The sin and cos functions . . . . . . . . . . . . . . . . . . . . . . 22.2 The tan function . . . . . . . . . . . . . . . . . . . . . . . . . . . 42.3 Converting from Polar to Cartesian coordinates . . . . . . . . . . 4

3 Relating cos, sin and tan with triangles 53.1 The good triangles . . . . . . . . . . . . . . . . . . . . . . . . . . 53.2 Rotations are irrelevant . . . . . . . . . . . . . . . . . . . . . . . 73.3 The sin, cos, tan and obtuse angles . . . . . . . . . . . . . . . . . 83.4 Special values of cos, sin and tan . . . . . . . . . . . . . . . . . . 8

4 Trigonometric Relations in Triangles 114.1 Law of Sines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 114.2 Law of Cosines . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12

5 References and Further Reading 13

6 Solutions 14

1 Introduction

Previously we saw we can assign to a point P in the plane the Cartesian coor-dinates �x, y� and the Polar coordinates �r, θ�, we represent now this map bythe downward arrows in the following diagram:

P

�x, y� �r, θ�

1

The fact that we can assign two types of coordinates to the same point Phas a very important consequence: the Cartesian coordinates must be relatedto the Polar coordinates; we represent this relation by the horizontal arrow.

Or goal is to explore this connection and how it is applied when speakingabout triangles. We start in Sec. 2 to introduce the trigonometric functionsfrom a geometrical point of view and then interpret them in triangles in Sec. 3.We finally end in Sec. 4 with trig identities that relate the side length of thetriangles with their internal angles.

2 Trigonometric functions

We seek to make clear the map between the point’s ”name” in polar coordinates�r, θ� and its unique ”name” in Cartesian coordinates �x, y�.We will focus first on the particular case of the point �1, θ� which belong

to the circumference of radius 1; the general mapping �r, θ�À �x, y� will beintroduced in due time.

The point �1, θ� must have a name in Cartesian coordinates, lets call it �x, y�.The map �1, θ�À �x, y� can be broken into two functions: the function thatassociates �1, θ� the value x, let’s call it the cosine function, abbreviated cos;the function that assign the y value the sine function, abbreviated sin.

2.1 The sin and cos functions

The domain, target and mapping procedure of these functions are specified inthe following definition.

Definition 2.1. Consider the unit circle S centered at the origin of a orthogonalreference frame. Let �x, y� be the Cartesian coordinates of a point P on S and θthe angle measurement in radians starting from the positive horizontal axis. Thefunction that converts the polar coordinates �1, θ� into the Cartesian coordinatey is defined as

sin � Rº ��1, 1�θÀ sin θ

where

sin θ �

~����������1�Draw a ref. frame and S.

2�Measure θ along the arc and mark P " S

3�Identify the y coordinate.

The function that converts �1, θ� into the x coordinate is:

cos � Rº ��1, 1�θÀ cos θ

2

where

cos θ �

~����������1�Draw a ref. frame and S.

2�Measure θ along the arc and mark P " S

3�Identify the x coordinate.

The procedures cos and sin are better understood pictorially, see Fig. 1.The first step in to simply draw the reference frame centered at the vertex ofthe angle. For the second step, recall from previous lecture that in the specialcase of circle of radius 1, S, the dimensionless number θ has the same numericalvalue as the distance s along the arc. By knowing s we know where to mark P ,the yellow dot in the unit circle S.

Figure 1: What cos and sin do to θ is to identify the x and y coordinates of ayellow point in the unit circle, these are represented by the horizontal andvertical blue lines. θ measures the ”cloud monster mouth” given by the dottedlines.

Once we have the point P we are ready to draw the horizontal and verticalline segments in blue, their lengths should be measured with respect to thereference frame. In Fig. 1 the left panel show the case where both these lengthare positive while in the right panel they are both negative.

Commentary: In general a function associates values elements of a set with anotherset, this association can be manifested for example by a table or by explicitly givingthe recipe one have to follow to realize the connection. For example when you sayf�x� � x

2� 1 what you mean is

f�x� � w1.Compute x times x.

2.Add 1(1)

The analytically expression x2� 1 is just a short hand for these two steps.

The trigonometric functions sin and cos also have a recipe to follow, which from

a geometrical point of view is given in the Definition 2.1, there is no analytically

expressions for these steps.

3

Exercise 1:

Compute cos and sin for the following angles θ: (a) 0, �2π and �4π; (b) �π, �3π

and �9π; (c) �π©2, �3π©2 and �13π©2; (d) �π©4, 9π©4 and 25π©4. Remember that

cos θ or sin θ means to follow the three step recipe in the definition applied to θ, this

is an exercise in geometry! It might be useful to start by decomposing the angle as a

integer number of full turns n2π plus a bit, δ. Mathematically speaking: θ � n2π � δ.

2.2 The tan function

From the construction of Fig. 1, we can also define the slope of the line segmentstarting at the origin and ending at the point P , as usual

slope �yx (2)

that ratio of the length of the blue line segments.This motivate us to define a new trigonometric function! The function that

assigns to each angle θ the slope of the corresponding right triangle

tan � R̃º R

θÀ tan θ �sin θ

cos θ

Notice that the domain is given by this special set R̃ that means all real numbersexcept numbers falling at r�π

2� n2π ¶ n " Zx � �

π̄2

; we can easily see why wemust exclude all these points, observing the denominator we find cos θ and thisfunction is zero precisely at all these points.

Exercise 2:

(a) Use the results from previous exercise to compute tan θ. (b) What can you say

about the set r�x, y� ¶ y � x tan θx?2.3 Converting from Polar to Cartesian coordinates

Now that we have a better understanding about the behavior of the trig func-tions we can easily make explicit the translation of a point name given in polarcoordinates into Cartesian coordinates.

The Definition 2.1 entails that a point P in the unit circle with polar coor-dinates �1, θ� has Cartesian coordinates �x, y�. The translation is hence

x � cos θ

y � sin θ(3)

Generalizing, a point P in a circle of radius r and Polar coordinates �r, θ� canbe translated into Cartesian coordinates �x, y� by

x � r cos θ

y � r sin θ(4)

4

The justification for this map is as follows: first note that any point P withcoordinates �x, y� the distance r between the origin and P obeys the equationr2� x

2� y

2. Now dividing by r

2yields

�xr 2 � �yr 2 � 1 (5)

This means that x©r and y©r are the coordinates of the point in the unit circle,thus

wxr � cos θy

r� sin θ

¼ wx � r cos θ

y � r sin θ(6)

While cos θ and sin θ have a target given by ��1, 1�, the functions r cos θ andr sin θ have ��r, r�.

It is possible to construct a conversion from Cartesian to Polar coordinates,but for that we have to wait for the next lecture.

Exercise 3:

If the point P has Polar coordinates �2, 3π©4� what are its Cartesian coordinates.

Make a picture.

3 Relating cos, sin and tan with triangles

Observing Fig. 1 we realize that the horizontal and vertical lines we drew inblue actually form a right triangle whose hypotenuse is the radius of the circle.The ratio of the horizontal length x with the hypotenuse yields the value ofcos θ, doing the same with the vertical line yield sin θ. Clearly the coordinates�x, y� of P describe the sides of this triangle and thus one can speak of thesefunctions as relating the angle measure with the sides of the triangle so as longas they are the sides of a right triangle.

There are three important aspects when dealing with triangles and trigono-metric functions.

3.1 The good triangles

Any triangle with the same ratio x©r or y©r will have the same cos θ or sin θ asillustrated in Fig. 2.

5

Figure 2: Three triangles with hypotenuse r0 smaller than 1, with radius 1 andanother with r2 larger than 1. cos θ and sin θ can be computed from ratios ofany triangle.

Hence we find

cos θ �x0r �

x11�x2r � � � � �

xr

sin θ �y0r �

y11�y2r � � � � �

yr

(7)

It doesn’t matter the triangle we draw to evaluate the ratios.The tan can also be evaluated for any of these triangles thus

tan θ �y1x1�y21�y3x3� � � � �

yx (8)

as we see in the Fig. 3

Figure 3: tan θ is the same for all these angles because the ratio y©x is alwaysthe same.

Since these trigonometric functions can be evaluated at any triangle sub-tended by the angle we can choose as we wish. However, some triangles exposesbetter the meaning of these function than other, these are the good triangles

6

to choose. Close inspection of equations (7) and (8) reveals that when thedenominator is 1 the evaluation is simpler and meaningful.

The good triangle to choose for cos and sin have hypothenuse mesuring 1.In this case two thing happen: the cos and sin pick the x and y components ofP , additionally θ is numerically the same as s as we saw in the previous lecture.

The tan function have different taste and prefers the triangle base equal to1, this makes tan θ coincide with the height of the yellow vertical line in Fig. 8.

3.2 Rotations are irrelevant

Since the angles of rotated triangles are congruent, their measure θ is the sameand thus the sin and cos of θ is always the same. For example, in Fig. 4 we findthree congruent triangles.

Figure 4: cos and sin of the internal angles are the same for all these righttriangles because they are congruent.

The value of cos, sin and tan of θ, β and π©2 is the same for the three. Thisproperty allow us to choose any of the versions to do the calculations (7) and(8), in particular we can do them in the standard unit circle angle as in Fig. 1or the left triangle in Fig. 4.

Exercise 4:

(a) Consider Fig. 2 and tan θ � 1. What can you say about cos θ and sin θ?

(b) For each triangle in Fig. 4 draw the unit circle centered at θ vertex and x�axis

aligned with the x side. Do the same for the β vertex but with the x�axis along y

side. Redo but now the x � axis is along the r side. There are many possibilities but

realize that each time you make one of these drawings you are essentially introducing

a ”mathematical protractor” to measure each angle. Realize what you have just done

is equivalent to just draw a single ”mathematical protractor” and then conveniently

rotate each of triangles to be ready for measure. Your rules and your conventions,

many possibilities!

7

3.3 The sin, cos, tan and obtuse angles

If we have at hands an obtuse triangle we have to construct a right triangle. InFig. 5 we drew one of the several possibilities

1for right triangles

Figure 5: Obtuse triangles inside a right triangles, the trig function evaluatedat θ give the same result regardless of the orientation

It is from these right triangles and not the obtuse ones that we compute theratios (7) and (8).

Exercise 5:

Let α be the obtuse angle measure in Fig. 5. What is the geometrical meaning of

cosα, sinα and tanα?

3.4 Special values of cos, sin and tan

Despite having the procedure we must resort to numerical calculations exceptfor some special angles.

Recall from the previous lecture the three special triangles, they are com-posed by the special angles, angles where we can actually do calculations byhand.

On Fig. 6 we find the two triangles from the previous lecture that we willuse, the equilateral triangle on the left and an almost equilateral triangle witha right triangle, lets call it a special name - the equiright triangle!

On the left panel of Fig. 6 we find our equilateral triangle divided in twohalves with α vertex centered at the origin; by the Pythagorean theorem we

find the base length asÔa2 � a2©2 � a

Ó3©2; since α � π©3 rad the angle

θ � α©2 � π©6 rad. Armed with our Trigonometric Definition 2.1 we can saythat

cosπ

6�

Ó3

2a �

1a �

Ó3

2

sinπ

6�a

2�

1a �

1

2

tanπ

6�

Ó3

3

(9)

1For other choices get inspired in Fig. 2

8

Figure 6: Special triangles. Left panel - Equilateral where the α is split intotwo halves, notice that if θ is positive then, the second has �θ. Right panel -the equiright with α equals θ.

From now on we know the values of these functions for θ � π©6 rad for anyright triangle irrespective of the rotation (cf. Sec. 3.2 ).

Observing now the right panel of Fig. 6 we find θ � π©4 rad and b �Ó

2a.Hence

cosπ

4�

aÓ2a�

Ó2

2

sinπ

4�

aÓ2a�

Ó2

2

tanπ

4� 1

(10)

From now on we know that these functions map θ � π©4 rad � 45`

into thesame number

Ó2©2 in any right triangle irrespective of rotations.

We can extract more information from the left panel of Fig. 6, imagine werotate the triangle so that its vertical side is now horizontal and the γ vertexcoincides with the origin, recall from previous lecture all angles in an equilateraltriangle have the same size, thus γ � π©3 rad, using our trig functions we arriveat

cosπ

3�a

2�

1a �

1

2

sinπ

3�

Ó3

2a �

1a �

Ó3

2

tanπ

3�

Ó3

(11)

9

Comparing with (9), we double the angle and permuted the outputs!The are two other special triangles that allow us to compute two other

mappings of these trigonometric functions. The ”triangle” with no height andthe ”triangle” with no base. Since these are extreme cases we illustrate themas extremes of a sequence of triangle

Figure 7: Sequence of right triangles with the hypotenuse of 1 and0 $ θ $ π©2. The yellow triangles are the extreme triangle since θ is close to 0and π©2, their base and height allow us to compute the cos and sin.

The yellow triangles in Fig. 7 tells us something important, the one with(almost) no height on the left says

cos 0 � 1

sin 0 � 0

tan 0 � 0

(12)

and the one with (almost) no base on the right says

cosπ

2� 0

sinπ

2� 1

(13)

As we approach the triangle on the right we note the slope rises without boundwe express this saying the tan goes to �.

We completed the most important mappings of the trigonometric functions,many other can now be reduced to this ones as we shall see. The following tableorganizes our results:

θ 0 π©6 π©4 π©3 π©2

sin 0 1©2Ó

2©2Ó

3©2 1

cos 1Ó

3©2Ó

2©2 1©2 0

tan 0Ó

3©3 1Ó

3 �

Table 1: Special outputs of ourtrigonometric functions, all derivedfrom the equilateral triangle, theequiright triangle and the twoextreme cases. If you know how you’llnever need to memorize this table.

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Exercise 6:

Compute: (a) cos�2π� � cos��4π� � sin��π©3�; (b) sin�3π©2� � cos��5π©3�; (c)

cos�37π©6� � sin��7π©4� � tan�13π©3�; (d) Eve wants to checks the answers of the

exam with you. She tells you that her calculation of the cosine of the angle gives her

1.01. What do you tell her?

4 Trigonometric Relations in Triangles

The following formulas relate internal angles of a triangle with the lengths itssides.

4.1 Law of Sines

Draw any triangle whose angles are α, β and γ whose opposite side lengths area, b and c, drop all three vertical lines starting at each vertex and perpendicularto the opposite side whose lengths are ha, hb and hc

Figure 8: Any triangle has its internal angles related to its side lengths. Thedotted lines are just a means to expose the relations.

From Fig. 8 and equation (7) we can write the following six equations

hb � c sinα � a sin γ

ha � b sin γ � c sinβ

hc � b sinα � a sinβ

(14)

Rearranging each second equation we know

c

sin γ�

a

sinα

b

sinβ�

c

sin γ

b

sinβ�

a

sinα

(15)

which in turn entails the remarkable Law of Sines:

a

sinα�

b

sinβ�

c

sin γ(16)

11

Exercise 7:

Consider the triangle in Fig. 8. Givens: b � 3, α � 70 deg, γ � 55 deg. Goal:

Compute a and c.

4.2 Law of Cosines

For any triangle, for example see Fig. 8 the following holds

a2� b

2� c

2� 2bc cosα (17)

It allow us to compute the length a from the length of the two other sides b andc at the price of knowing the angle α. In particular if α � π©2 we have a righttriangle, then the law of cosines reduces to a very famous formula (Who is she?)

The justification of this relation can be given in many ways, the simplest ofwhich uses vector notation, but since we haven’t yet arrive there, we will usebasic arithmetic.

Observing Fig. 8 we know that hb splits the triangle into two right trianglesand b in two parts, one is c cosα and the other is whatever it’s left, b � c cosα.Using the Pythagorean theorem on each of the two right triangles we know

h2b � c

2� �c cosα�2 � a2 � �b � c cosα�2 (18)

Focusing now on the second equation we expand the polynomial and isolate a2

to the lhs, from this we find (17).It should be clear that we can repeat this argument for any other angles

yielding the two other Laws of Cosines for a triangle in Fig. 8

b2� a

2� c

2� 2ac cosβ

c2� a

2� b

2� 2ab cos γ

(19)

Exercise 8:

(a) Consider again the triangle in Fig. 8. Givens: c � 10, b � 12 α � 30 deg. Goal:

Compute a. (b) You and Eve are given the task of carefully draw a triangle with sides

1, 2 and 3. Give an example of such a triangle. To check your answer recall that the

Law of Cosines must be true for all triangles. Is it true for the triangle you drew?

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5 References and Further Reading

Basic and with many exercises:

� I.M. Gelfand and Mark Saul, Trigonometry

� Carl Stitz and Jeff Zeager, College Trigonometry

Online Lectures and notes from PART III: THE CIRCULAR FUNC-TIONS by Herbert I. Gross

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6 Solutions

Solution 1:

To decompose for example 13π©2, means to write it in the form 13π©2 �n2π � δ for some integer n and angle δ we must find. We do it by trial anderror. In this example 13π©2 � 3 � 2π � π©2. Decomposing all angles we find

For n " Z we have: (a) cos�0�n2π� � 1 and sin�0�n2π� � 0; (b) cos��π�n2π� � �1 and sin��π � n2π� � 0; (c) cos��π©2 � n2π� � 0 and sin��π©2 �n2π� � �1; (d) cos��π©4 � n2π� � Ó

2©2 and sin��π©4 � n2π� � �Ó2©2

Solution 2:

(a) tan�0 � n2π� � 0; tan��π � n2π� � 0; tan��π©2 � n2π� is not definedsince it goes to �; tan��π©4 � n2π� � 1.

(b) Its the set of Cartesian coordinates of all point constituting a line thatpasses through the origin and has slope tan θ ie. makes an angle θ with the hor-izontal axis of your reference frame. Appreciate the fact that the elementhoodtest has the form y � x �m.

Solution 3:

Decompose the angle as 3π©4 � π©2 � π©4 and compute

�2 cos�π©2 � π©4�, 2 sin�π©2 � π©4�� � ��Ó2,Ó

2�

Solution 4:

(a)

1 � tan θ �sin θ

cos θ

This implies that cos θ � sin θ, that is x � y. For a triangle with hypotenuse rand sides x and y, we have thus r

2� x

2� y

2� 2x

2� 2y

2. Solving for x (or y)

we find x � y � rÓ

2©2.

14

(b) We give just one example:

Figure 9: Notice the choice of ”mathematical protractor” yields β positive (cwis positive) but θ negative (ccw is positive). Other choices can make both ��,�� or ��.

Solution 5:

We conveniently introduce the following the circle with radius R and refer-ence frame aligned with the base of the triangle.

Figure 10: cosα � x¬

R� r cos θ � x. sinα � y

R� r sin θ; tanα � y

x¬�

r sin θr cos θ�x

.

15

Solution 6:

(a) 2 �Ó32

; (b) � 12; (c)

Õ32�

Ó32

; (d) Eve: I’m just testing you ;)

Solution 7:

a � b sinαsin�π�α�γ� ; c � b

sinγ

sin�π�α�γ�

Solution 8:

(a) a2� b

2� c

2� 2bc cosα� 2π

360(b) 1

2� 2

2� 3

2� 2 � 2 � 3 cosα ¼ cosα �

�1 ¼ α � π. This means we have no triangle! It would be useful for youto check that for any triangle to exist the sum of any two sides must be largerthan the third !

16