em03

ELEMENTARY MATHEMATICS

W W L CHEN and X T DUONG

c W W L Chen, X T Duong and Macquarie University, 1999.This work is available free, in the hope that it will be useful.

Any part of this work may be reproduced or transmitted in any form or by any means, electronic or mechanical, including

photocopying, recording, or any information storage and retrieval system, with or without permission from the authors.

Chapter 3

TRIGONOMETRY

3.1. Radian and Arc Length

The number plays a central role in the study of trigonometry. We all know that a circle of radius 1has area and circumference 2. It is also very useful in describing angles, as we shall now show.

Let us split a circle of radius 1 along a diameter into two semicircles as shown in the picture below.

The circumference of the circle is now split into two equal parts, each of length and each suntendingan angle 180. If we use the convention that = 180, then the arc of the semicircle of radius 1 will bethe same as the angle it subtends. If we further split the arc of the semicircle of radius 1 into two equalparts, then each of the two parts forms an arc of length /2 and subtends an angle 90 = /2. In fact,any arc of a circle of radius 1 which subtends an angle must have length under our convention.

We now formalize our discussion so far.

Definition. An angle of 1 radian is dened to be the angle subtended by an arc of length 1 on a circleof radius 1.

This chapter was written at Macquarie University in 1999.

r

r

x

y

1

32 W W L Chen and X T Duong : Elementary Mathematics

Remarks. (1) Very often, the term radian is omitted when we discuss angles. We simply refer to anangle 1 or an angle , rather than an angle of 1 radian or an angle of radian.

(2) Simple calculation shows that 1 radian is equal to (180/) = 57.2957795 . . .. Similarly, wecan show that 1 is equal to (/180) radian = 0.01745329 . . . radian. In fact, since is irrational, thedigits do not terminate or repeat.

(3) We observe the following special values:

6= 30,

4= 45,

3= 60,

2= 90, = 180, 2 = 360.

Consider now a circle of radius r and an angle given in radian, as shown in the picture below.

Clearly the length s of the arc which subtends the angle satises s = r, while the area A of the sectorsatises

A = r2 2

=12r2.

Note that r2 is equal to the area inside the circle, while /2 is the proportion of the area in question.

3.2. The Trigonometric Functions

Consider the xy-plane, together with a circle of radius 1 and centred at the origin (0, 0). Suppose that is an angle measured anticlockwise from the positive x-axis, and the point (x, y) on the circle is asshown in the picture below.

We denecos = x and sin = y.

Furthermore, we dene

tan =sin cos

=y

x, cot =

cos sin

=x

y, sec =

1cos

=1x

and csc =1

sin =

1y.

2 -1

1

34

Chapter 3 : Trigonometry 33

Remarks. (1) Note that tan and sec are dened only when cos = 0, and that cot and csc aredened only when sin = 0.

(2) It is a good habit to always measure an angle from the positive x-axis, using the convention thatpositive angles are measured anticlockwise and negative angles are measured clockwise, as illustratedbelow.

(3) We then observe that

cos =horizontal sidehypothenuse

and sin =vertical sidehypothenuse

,

as well as

tan =vertical side

horizontal sideand cot =

horizontal sidevertical side

,

with the convention that

horizontal side{

> 0 (to the right of the (vertical) y-axis),< 0 (to the left of the (vertical) y-axis),

and

vertical side{

> 0 (above the (horizontal) x-axis),< 0 (below the (horizontal) x-axis),

whilehypothenuse > 0 (always).

(4) It is also useful to remember the CAST rule concerning sine, cosine and tangent.

S (sin > 0) A (all > 0)

T (tan > 0) C (cos > 0)

PYTHAGOREAN IDENTITIES. For every value R for which the trigonometric functions inquestion are dened, we have(a) cos2 + sin2 = 1;(b) 1 + tan2 = sec2 ; and(c) 1 + cot2 = csc2 .

1

-1

2-2

y

x-

1

-1

2-2

y

x-

2-2 -

y

x


Proof. Note that (a) follows from the classical Pythagorass theorem. If cos = 0, then dividing bothsides of (a) by cos2 gives (b). If sin = 0, then dividing both sides of (a) by sin2 gives (c).

We sketch the graphs of the functions y = sinx, y = cosx and y = tanx for 2 x 2 below:

y = sinx

y = cosx

y = tanx

We shall use these graphs to make some observations about trigonometric functions.


PROPERTIES OF TRIGONOMETRIC FUNCTIONS.(a) The functions sinx and cosx are periodic with period 2. More precisely, for every x R, we have

sin(x + 2) = sinx and cos(x + 2) = cosx.

(b) The functions tanx and cotx are periodic with period . More precisely, for every x R for whichthe trigonometric function in question is dened, we have

tan(x + ) = tanx and cot(x + ) = cotx.

(c) The function sinx is an odd function, and the function cosx is an even function. More precisely,for every x R, we have

sin(x) = sinx and cos(x) = cosx.

(d) The functions tanx and cotx are odd functions. More precisely, for every x R for which thetrigonometric function in question is dened, we have

tan(x) = tanx and cot(x) = cotx.

(e) For every x R, we have

sin(x + ) = sinx and cos(x + ) = cosx.

(f) For every x R, we have

sin( x) = sinx and cos( x) = cosx.

(g) For every x R for which the trigonometric function in question is dened, we have

tan( x) = tanx and cot( x) = cotx.

(h) For every x R, we have

sin(x +

2

)= cosx and cos

(x +

2

)= sinx.

(i) For every x R, we have

sin(

2 x

)= cosx and cos

(2 x

)= sinx.

(j) For every x R for which the trigonometric functions in question are dened, we have

tan(x +

2

)= cotx and cot

(x +

2

)= tanx.

(k) For every x R for which the trigonometric function in question is dened, we have

tan(

2 x

)= cotx and cot

(2 x

)= tanx.

Remark. There is absolutely no need to remember any of these properties! We shall discuss latersome trigonometric identities which will give all the above as special cases.

/6

1

-/3

1


Example 3.2.1. Consider the following picture.

Clearly the triangle shown is an equilateral triangle, with all three sides of equal length. It is also clearthat sin(/6) is half the length of the vertical side. It follows that we must have sin(/6) = 1/2. To ndthe precise value for cos(/6), we rst observe that cos(/6) > 0. On the other hand, it follows from therst of the Pythagorean identities that cos2(/6) = 3/4. Hence cos(/6) =

3/2. We can then deduce

that tan(/6) = 1/

3.

Example 3.2.2. We have tan(13/6) = tan(7/6) = tan(/6) = 1/

3. Note that we have used part(b) of the Properties of trigonometric functions, as well as the result from Example 3.2.1.


Clearly the triangle shown is an equilateral triangle, with all three sides of equal length. It is also clearthat cos(/3) is half the length the horizontal side. It follows that we must have cos(/3) = 1/2.To nd the precise value for sin(/3), we rst observe that sin(/3) < 0. On the other hand,it follows from the rst of the Pythagorean identities that sin2(/3) = 3/4. Hence sin(/3) =3/2. Alternatively, we can deduce from Example 3.2.1 by using parts (c) and (i) of the Properties of

1

5/4

1

7/4

1


trigonometric functions that

cos(

3

)= cos

3= sin

(2

3

)= sin

6=

12

and

sin(

3

)= sin

3= cos

(2

3

)= cos

6=

3

2.

We can then deduce that tan(/3) = 3.


Clearly the triangle shown is a right-angled triangle with the two shorter sides of equal length andhypothenuse of length 1. It is also clear that sin(5/4) is the length of the vertical side, with a signattached as it is below the horizontal axis. On the other hand, it is also clear that cos(5/4) is thelength of the horizontal side, again with a sign attached as it is left of the vertical axis. It followsfrom Pythagorass theorem that sin(5/4) = cos(5/4) = y, where y < 0 and

y2 + y2 = 1. Clearly

y = 1/2. We can then deduce that tan(5/4) = 1.

Example 3.2.5. To nd all the solutions of the equation sinx = 1/2 in the interval 0 x < 2,we consider the following picture.

1 1

1

-/3

2/3

1

1

4/3


Using Pythagorass theorem, it is easy to see that the two triangles shown both have horizontal side oflength 1/

2, the same as the length of their vertical sides. Clearly x = 5/4 or x = 7/4.

Example 3.2.6. Suppose that we wish to nd all the solutions of the equation cos2 x = 1/4 in theinterval /2 < x . Observe rst of all that either cosx = 1/2 or cosx = 1/2. We consider thefollowing picture.

Using Pythagorass theorem, it is easy to see that the three triangles shown both have vertical side oflength

3/2. Clearly x = /3, x = /3 or x = 2/3.

Example 3.2.7. Convince yourself that the only two solutions of the equation cos2 x = 1 in the interval0 x < 2 are x = 0 and x = .

Example 3.2.8. To nd all the solutions of the equation tanx =

3 in the interval 0 x < 2, weconsider the following picture.

Since tanx > 0, it follows from the CAST rule that we can restrict our attention to the rst and thirdquadrants. It is easy to check that the two triangles shown have horizontal sides of length 1/2 andvertical sides of length

3/2. Clearly x = /3 or x = 4/3.

11

1

7/6


Example 3.2.9. Suppose that we wish to nd all the solutions of the equation tan2 x = 1/3 in theinterval 0 x 3/2. Observe rst of all that either tanx = 1/3 or tanx = 1/3. We consider thefollowing picture.

It is easy to check that the three triangles shown have horizontal sides of length

3/2 and vertical sidesof length 1/2. Clearly x = /6, x = 5/6 or x = 7/6.

Example 3.2.10. Convince yourself that the only two solutions of the equation secx =

2 in theinterval 0 x < 2 are x = /4 and x = 7/4.

Example 3.2.11. For every x R, we have

sin3 x + sinx cos2 x = sinx sin2 x + sinx cos2 x = (sinx)(sin2 x + cos2 x) = sinx,

in view of the rst of the Pythagorean identities.

Example 3.2.12. For every x R such that cosx = 0, we have

(secx tanx)(secx + tanx) = sec2 x tan2 x = 1,

in view of the second of the Pythagorean identities.

Example 3.2.13. For every x R such that cosx = 1, we have1

1 cosx +1

1 + cosx=

(1 + cosx) + (1 cosx)(1 cosx)(1 + cosx) =

21 cos2 x =

2sin2 x

= 2 csc2 x.


(cosx + sinx)2 + (cosx sinx)2 = (cos2 x + 2 cosx sinx + sin2 x) + (cos2 x 2 cosx sinx + sin2 x)= (1 + 2 cosx sinx) + (1 2 cosx sinx) = 2.

Example 3.2.15. For every x R such that the expression on the left hand side makes sense, we have1 + cotx

cscx 1 + tanx

secx= (1 + cotx) sinx (1 + tanx) cosx =

(1 +

cosxsinx

)sinx

(1 +

sinxcosx

)cosx

= (sinx + cosx) (cosx + sinx) = 0.


Example 3.2.16. Let us return to Example 3.2.5 where we showed that the solutions of the equationsinx = 1/2 in the interval 0 x < 2 are given by x = 5/4 and x = 7/4. Suppose now that wewish to nd all the values x R that satisfy the same equation. To do this, we can use part (a) of theProperties of trigonometric functions, and conclude that the solutions are given by

x =54

+ 2k or x =74

+ 2k,

where k Z.

Example 3.2.17. Consider the equation sec(x/2) = 2. Then cos(x/2) = 1/2. If we rst restrict ourattention to 0 x/2 < 2, then it is not dicult to see that the solutions are given by x/2 = /3 andx/2 = 5/3. Using part (a) of the Properties of trigonometric functions, we conclude that without therestriction 0 x/2 < 2, the solutions are given by

x

2=

3+ 2k or

x

2=

53

+ 2k,

where k Z. It follows that

x =23

+ 4k or x =103

+ 4k,

where k Z.

3.3. Some Trigonometric Identities

Consider a triangle with side lengths and angles as shown in the picture below:

c

b

J

J

J

J

J

J

J

J

J

J

J

J

J

J

J

J

J

aB C

A

J

J

J

J

J

J

J

J

J

J

J

J

J

J

J

J

J

SINE RULE. We havea

sinA=

b

sinB=

c

sinC.

COSINE RULE. We have

a2 = b2 + c2 2bc cosA,b2 = a2 + c2 2ac cosB,c2 = a2 + b2 2ab cosC.

Sketch of Proof. Consider the picture below.

c

b

J

J

J

J

J

J

J

J

J

J

J

J

J

J

J

J

JB C

J

J

J

J

J

J

J

J

J

J

J

J

J

J

J

J

J

oo

a


Clearly the length of the vertical line segment is given by c sinB = b sinC, so that

b

sinB=

c

sinC.

This gives the sine rule. Next, note that the horizontal side of the the right-angled triangle on the lefthas length c cosB. It follows that the horizontal side of the right-angled triangle on the right has lengtha c cosB. If we now apply Pythagorass theorem to this latter triangle, then we have

(a c cosB)2 + (c sinB)2 = b2,

so thata2 2ac cosB + c2 cos2 B + c2 sin2 B = b2,

whenceb2 = a2 + c2 2ac cosB.

This gives the cosine rule.

We mentioned earlier that there is no need to remember any of the Properties of trigonometricfunctions discussed in the last section. The reason is that they can all be deduced easily from theidentities below.

SUM AND DIFFERENCE IDENTITIES. For every A,B R, we have

sin(A + B) = sinA cosB + cosA sinB and sin(AB) = sinA cosB cosA sinB,

as well as

cos(A + B) = cosA cosB sinA sinB and cos(AB) = cosA cosB + sinA sinB.

Remarks. (1) Proofs can be sketched for these identities by drawing suitable pictures, although suchpictures are fairly complicated. We omit the proofs here.

(2) It is not dicult to remember these identities. Observe the pattern

sin = sin cos cos sin and cos = cos cos sin sin .

For small positive angles, increasing the angle increases the sine (thus keeping signs) and decreases thecosine (thus reversing signs).

(3) One can also deduce analogous identities for tangent and cotangent. We have

tan(A + B) =sin(A + B)cos(A + B)

=sinA cosB + cosA sinBcosA cosB sinA sinB .

Dividing both the numerator and denominator by cosA cosB, we obtain

tan(A + B) =

sinAcosA

+sinBcosB

1 sinA sinBcosA cosB

=tanA + tanB

1 tanA tanB .

Similarly, one can deduce that

tan(AB) = tanA tanB1 + tanA tanB

.



sin(

2 x

)= sin

2cosx cos

2sinx = cosx,

andcos

(2 x

)= cos

2cosx + sin

2sinx = sinx.

These form part (i) of the Properties of trigonometric functions.

Of particular interest is the special case when A = B.

DOUBLE ANGLE IDENTITIES. For every x R, we have(a) sin 2x = 2 sinx cosx; and(b) cos 2x = cos2 x sin2 x = 1 2 sin2 x = 2 cos2 x 1.

HALF ANGLE IDENTITIES. For every y R, we have

sin2(y

2

)=

1 cos y2

and cos2(y

2

)=

1 + cos y2

.

Proof. Let x = y/2. Then part (b) of the Double angle identities give

cos y = 1 2 sin2(y

2

)= 2 cos2

(y2

) 1.

The results follow easily.

Example 3.3.2. Suppose that we wish to nd the precise values of cos(3/8) and sin(3/8). Wehave

cos2(3

8

)=

1 + cos(3/4)2

.

It is not dicult to show that cos(3/4) = 1/2, so that

cos2(3

8

)=

12

(1 1

2

)=

2 12

2.

It is easy to see that cos(3/8) > 0, and so

cos(3

8

)=

2 12

2.

Similarly, we have

sin2(3

8

)=

1 cos(3/4)2

=12

(1 +

12

)=

2 + 12

2.

It is easy to see that sin(3/8) < 0, and so

sin(3

8

)=

2 + 12

2.


Example 3.3.3. Suppose that is an angle in the rst quadrant and is an angle in the thirdquadrant. Suppose further that sin = 3/5 and cos = 5/13. Using the rst of the Pythagoreanidentities, we have

cos2 = 1 sin2 = 1625

and sin2 = 1 cos2 = 144169

.

On the other hand, using the CAST rule, we have cos > 0 and sin < 0. It follows that cos = 4/5and sin = 12/13. Then

sin( ) = sin cos cos sin = 35

( 5

13

) 4

5

(12

13

)=

3365

and

cos( ) = cos cos + sin sin = 45

( 5

13

)+

35

(12

13

)= 56

65,

so that

tan( ) = sin( )cos( ) =

3356

.


sin 6x cos 2x cos 6x sin 2x = sin(6x 2x) = sin 4x = 2 sin 2x cos 2x.

Note that the rst step uses a dierence identity, while the last step uses a double angle identity.

Example 3.3.5. For appropriate values of , R, we have

sin( + )sin( ) =

sin cos + cos sinsin cos cos sin =

1 + cot tan1 cot tan .

Note that the rst step uses sum and dierence identities, while the second step involves dividing boththe numerator and the denominator by sin cos.

Example 3.3.6. For every x R for which sin 4x = 0, we have

cos 8xsin2 4x

=cos2 4x sin2 4x

sin2 4x= cot2 4x 1.

Note that the rst step involves a double angle identity.

Example 3.3.7. For every x R for which cosx = 0, we have

sin 2x1 sin2 x =

2 sinx cosxcos2 x

= 2 tanx.

Note that the rst step involves a double angle identity as well as a Pythagorean identity.

Example 3.3.8. For every x R for which sin 3x = 0 and cos 3x = 0, we have

cos2 3x sin2 3x2 sin 3x cos 3x

=cos 6xsin 6x

= cot 6x.

Note that the rst step involves double angle identities.


Example 3.3.9. This example is useful in calculus for nding the derivatives of the sine and cosinefunctions. For every x, h R with h = 0, we have

sin(x + h) sinxh

=sinx cosh + cosx sinh sinx

h= (cosx) sinh

h+ (sinx) cosh 1

hand

cos(x + h) cosxh

=cosx cosh sinx sinh cosx

h= (sinx) sinh

h+ (cosx) cosh 1

h.

When h is very close to 0, then

sinhh

1 and cosh 1h

0,so that

sin(x + h) sinxh

cosx and cos(x + h) cosxh

sinx.

This is how we show that the derivatives of sinx and cosx are respectively cosx and sinx.

Problems for Chapter 3

1. Find the precise value of each of the following quantities, showing every step of your argument:

a) sin43

b) tan43

c) cos34

d) tan34

e) tan654

f) sin(47

6

)g) cos

253

h) cot(37

2

)i) sin

53

j) tan53

k) cos(53

6

)l) cot

(53

6

)2. Find all solutions for each of the following equations in the intervals given, showing every step of

your argument:

a) sinx = 12, 0 x < 2 b) sinx = 1

2, 0 x < 4

c) sinx = 12, x < d) sinx = 1

2, 0 x < 3

e) cos2 x =34, 0 x < 2 f) cos2 x = 3

4, x < 3

g) tan2 x = 3, 0 x < 2 h) tan2 x = 3, 2 x < i) cot2 x = 3, 0 x < 2 j) sec2 x = 2, 0 x < 5

2

k) cosx = 12, 0 x < 2 l) cosx = 1

2, 0 x < 4

m) tan2 x = 1, 2 x < 4 n) sin2 x = 14,

2 x < 2

3. Simplify each of the following expressions, showing every step of yor argument:

a)sinx secx sin2 x tanx

sin 2xb) sin 3x cosx + sinx cos 3x

c) cos 5x cosx + sinx sin 5x d) sin 3x cos 2x sinx 2 sinx cos2 xe)

sin 4x(cos2 x sin2 x) sinx cosx f) cos 4x cos 3x 4 sinx sin 3x cosx cos 2x

g) 2 sin 3x cos 3x cos 5x (cos2 3x sin2 3x) sin 5x


4. We know that

sin

6=

12, cos

6=

32

, sin

4= cos

4=

12

and524

=12

(4

+

6

).

We also know that

cos( + ) = cos cos sin sin and cos 2 = 2 cos2 1 = 1 2 sin2 .

Use these to determine the exact values of

cos524

and sin524

.

[Hint: Put your calculators away. Your answers will be square roots of expressions involving thenumbers

2 and

3.]

5. Find the precise value of each of the following quantities, showing every step of your argument:

a) cos(

8

)b) sin

(

8

)c) cos

712

d) tan712

6. Use the sum and dierence identities for the sine and cosine functions to deduce each of the followingidentities:a) sin(x + ) = sinx b) cos(x) = cosx c) tan( x) = tanxd) cot

(2

+ x)

= tanx

em03

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