P1: FXS/ABE P2: FXS CHAPTER 11 - Cambridge University Press · 2008. 9. 22. · P1: FXS/ABE P2: FXS...

P1: FXS/ABE P2: FXS

9780521740494c11.xml CUAU033-EVANS September 9, 2008 13:42

C H A P T E R

11Circular functions II

ObjectivesTo further explore the symmetry properties of circular functions

To further understand and sketch the graphs of circular functions

To solve circular function equations

To evaluate simple trigonometric expressions using trigonometric identities

To prove simple trigonometric identities

To apply addition theorems for circular functions

To apply double angle formulas for circular functions

To simplify expressions of the form a cos x + b sin xTo sketch graphs of functions of the form f (x) = a cos x + b sin xTo solve equations of the form a cos x + b sin x = c

y

x

– θθ

θa

b

a

b

P(θ)π2

Pπ2

– θ

y

x

b

b

a

a

π2

+ θP(θ)

Pπ2

+ θ

θ

θ

11.1 Further symmetry propertiesComplementary relationships

sin(�

2− �

)= a

and since a = cos �sin

(�2

− �)

= cos �Similarly

cos(�

2− �

)= b

and since b = sin �cos

(�2

− �)

= sin �

sin(�

2+ �

)= a = cos �

cos(�

2+ �

)= −b = −sin �

297

SAM

PLE

Cambridge University Press • Uncorrected Sample Pages • 978-0-521-61252-4 2008 © Evans, Lipson, Jones, Avery, TI-Nspire & Casio ClassPad material prepared in collaboration with Jan Honnens & David Hibbard

P1: FXS/ABE P2: FXS


298 Essential Advanced General Mathematics

Example 1

If sin � = 0.3 and cos � = 0.8, find the values ofa sin

(�2

− �)

b cos(�

2+ �

)c sin(−�)

Solution

a sin(�

2− �

)= cos �= 0.8

b cos(�

2+ �

)= −sin �= −0.3

c sin (−�) = − sin �= −0.3

Exercise 11A

1 If sin x = 0.3, cos � = 0.6 and tan � = 0.7, find the values ofExample 1a cos(−�) b sin

(�2

+ �)

c tan(−�) d cos(�

2− x

)e sin(−x) f tan

(�2

− �)

g cos(�

2+ x

)h sin

(�2

− �)

i sin

(3�

2+ �

)j cos

(3�

2− x

)

11.2 Addition of ordinatesExample 2

Using the same scale and axes, sketch the graphs of y1 = 2 sin x and y2 = 3 cos 2x for0 ≤ x ≤ 2�.Use addition of ordinates to sketch the graph of y = 2 sin x + 3 cos 2x .

x

y

3210

–1–2–3

–5–4

π2

2ππ

y1 = 2sinx

y2 = 3cos2x

y = 2sinx + 3cos2x

3π2

Solution

The graphs of y1 = 2 sin x and y2 = 3 cos 2x are shown below.To obtain points on the graph of y = 2 sin x + 3 cos 2x the process of addition ofordinates is used.

Let y = y1 + y2 when y1 = 2 sin x and y2 = 3 cos 2xe.g., at

x = 0, y = 0 + 3 = 3x = �

4, y = 2√

2+ 0 = 2√

2= √2

x = �2

, y = 2 − 3 = −1x = �, y = 0 + 3 = 3x = 3�

2,y = −2 − 3 = −5

and so on.

SAM

PLE


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Chapter 11 — Circular functions II 299

Using the TI-NspireCheck that the calculator is in Radian

mode. Open a Graphs & Geometryapplication ( 2) and enter the

functions

f 1 (x) = 2 sin (x)f 2 (x) = 3 cos (2x)f 3 (x) = f 1 (x) + f 2 (x)

The graph of f (3) x is the heavier line.

Add Function Table (b 2 ) and splitthe screen as shown using the Tools menu(/ 5 23) to see that the values

of f 1 (x) and f 2 (x) add to give f 3 (x).

Use the down arrow ( ) to see more xvalues.

Using the Casio ClassPadSet the calculator to Radian mode.

Enter the functions y = sin (x) and y = 12

.

The window settings should be as shown. Tap

$ to produce the graph. Ensure the graph

window is selected (bold border) and tap Analysis,G-solve, intersect to find decimal approximationsfor the solutions. The scroll key moves the cursor

between solutions.SAM

PLE


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To find exact solutions, the window is

used.

Enter and highlight the equation sin (x) = 12

.

Tap Interactive, Equation/inequality, solve andensure the variable is set to x.

The answer returned is

x = 2�∗constn(9) + �6

,

x = 2�∗constn(10) + 5�6

.

This may be read as

x = 2m� + �6

, 2n� + 5�6

.

It should be clear that there are 4 solutions to the problem. Hence, the values for each of

m and n will be required which produce a solution in the domain. In this case the values are

m = 0, 1 and n = 0, 1. The solutions are

x = �6

,11�

6,

13�

6,

23�

6.

Exercise 11B

1 Use addition of ordinates to sketch the graphs ofExample 2

a y = 2 sin � + cos � b y = 3 cos 2� + 2 sin 2�c y = 1

2sin 2� − cos � d y = 3 sin � + cos 2�

e y = 4 sin � − 2 cos �

11.3 Sketch graphs of the tangent functionA table of values for y = tan x is given below. Use a calculator to check these values.

x −� −3�4

−�2

−�4

0�

4

�

2

3�

4�

5�

4

3�

2

7�

42�

9�

4

5�

2

11�

43�

y 0 1 undefined −1 0 1 undefined −1 0 1 undefined −1 0 1 undefined −1 0SAM

PLE


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The graph of y = tan x is given below.

x

y

–π1

10

2

2 3

2π 3π8 94 5 6 7 10

π2

3π2

–π2

5π2

–3–4 –2 –1–1

–2

π

Note: x = −�2

,�

2,

3�

2and

5�

2are asymptotes.

Observations from the graph1 The graph repeats itself every � units, i.e., the period of tan is �.

2 The range of tan is R.

Exercise 11C

1 Sketch the graph of each of the following, showing one complete cycle.

a y = tan 2x b y = 2 tan 3x c y = 2 tan(

x + �4

)d y = 3 tan x + 1 e y = 2 tan

(x + �

2

)+ 1 f y = 3 tan 2

(x − �

4

)− 2

11.4 General solution of circular function equationsThe solution of circular function equations has been discussed in Section 10.9 for functions

over a restricted domain. In this section, we consider the general solutions of such equations

over the maximal domain for each function.

If a circular function equation has one or more solutions in one ‘cycle’, then it will have

corresponding solutions in each ‘cycle’ of its domain, i.e., there will be an infinite number of

solutions.

For example, if cos x = a, then the solution in the interval [0, �] is given by:x = cos−1(a)

By the symmetry properties of the cosine function, other solutions are given by:

−cos−1(a), ±2� + cos−1(a), ±2� − cos−1(a), ±4� + cos−1(a),±4� − cos−1(a), . . . and so on.

In general, if cos (x) = a,x = 2n� ± cos−1(a), where n ∈ Z and a ∈ [−1, 1]

SAM

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Similarly, if tan (x) = a,x = n� + tan−1(a), where n ∈ Z and a ∈ R

If sin (x) = a,x = 2n� + sin−1(a) or x = (2n + 1)� − sin−1(a), where n ∈ Z and a ∈ [−1, 1]

Note: An alternative and more concise way to express the general solution of sin (x) = a is:x = n� + (−1)n sin−1(a), where n ∈ Z and a ∈ [−1, 1]

Example 3

Find the general solution to each of the following equations.

a cos (x) = 0.5 b √3 tan (3x) = 1 c 2 sin (x) = √2

Solution

a x = 2n� ± cos−1(0.5)= 2n� ± �

3

= (6n ± 1)�3

, n ∈ Z

b tan (3x) = 1√3

3x = n� + tan−1(

1√3

)

= n� + �6

= (6n + 1)�6

x = (6n + 1)�18

, n ∈ Zc sin (x) = 1√

2

x = 2n� + sin−1(

1√2

)or x = (2n + 1)� − sin−1

(1√2

)= 2n� + �

4= (2n + 1)� − �

4

= (8n + 1)�4

, n ∈ Z = (8n + 3)�4

, n ∈ Z

Using the TI-NspireCheck that the calculator is in Radian mode.

a Use Solve( ) from the Algebra menu(b 31) and complete as shown.

Note the use of1

2rather than 0.5 to

ensure that the answer is exact.SA

MPL

E


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b Complete as shown.

c Complete as shown.

Using the Casio ClassPada Enter and highlight the equation cos (x) = 0.5,

tap Interactive, Equation/inequation, solveand ensure the variable is set to x.

b Enter and highlight the equation

(3) tan (3x) = 1, tap Interactive,Equation/inequation, solve and ensure thevariable is set to x.

c Enter and highlight the equation (2) sin (x) = 1,tap Interactive, Equation/inequation, solveand ensure the variable is set to x.

Example 4

Find the first three positive solutions to each of the following equations.

a cos (x) = 0.5 b √3 tan (3x) = 1 c 2 sin (x) = √2

SAM

PLE


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Solution

a The general solution (from Example 3) is given by x = (6n ± 1)�3

, n ∈ Z

When n = 0, x = ±�3

, and when n = 1, x = 5�3

or x = 7�3

The first three positive solutions of cos (x) = 0.5 are x = �3

,5�

3,

7�

3

b The general solution (from Example 3) is given by x = (6n + 1)�18

, n ∈ ZWhen n = 0, x = �

18, and when n = 1, x = 7�

18, and when n = 2, x = 13�

18

The first three positive solutions of√

3 tan (3x) = 1 are x = �18

,7�

18,

13�

18

c The general solution (from Example 3) is given by x = (8n + 1)�4

or

x = (8n + 3)�4

, n ∈ Z

When n = 0, x = �4

or3�

4, and when n = 1, x = 9�

4or x = 11�

4

The first three positive solutions of 2 sin (x) = √2 are x = �4

,3�

4,

9�

4

Exercise 11D

1 Find the general solution to each of the following equations.Example 3

a sin (x) = 0.5 b 2 cos (3x) = √3 c √3 tan (x) = −3

2 Find the first two positive solutions to each of the following equations.Example 4

a sin (x) = 0.5 b 2 cos (3x) = √3 c √3 tan (x) = −3

3 Find the general solution to 2 cos(

2x + �4

)= √2, and hence find all the solutions for x in

the interval (−2�, 2�).

4 Find the general solution to√

3 tan(�

6− 3x

)− 1 = 0, and hence find all the solutions for

x in the interval [−�, 0].

5 Find the general solution to 2 sin (4�x) + √3 = 0, and hence find all the solutions for x inthe interval [−1, 1].

11.5 Trigonometric identitiesReciprocal functionsThe functions sin, cos, and tan can be used to form three other functions called the reciprocal

circular functions.

sec � = 1cos �

(cos � �= 0) cosec � = 1sin �

(sin � �= 0)

cot � = cos �sin �

(sin � �= 0)Note: For cos � �= 0 and sin � �= 0, cot � = 1

tan �and tan � = 1

cot �

SAM

PLE


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Example 5

Find the exact value of each of the following.

a sec2�

3b cot

5�

4c cosec

7�

4

Solution

a sec2�

3= 1

cos2�

3

= 1cos

(� − �

3

)= 1

−cos �3

= 1−1

2= −2

b cot5�

4=

cos5�

4

sin5�

4

=cos

(� + �

4

)sin

(� + �

4

)= −1√

2÷ −1√

2= 1

c cosec7�

4

= 1sin

(2� − �

4

)= 1

−sin(�

4

)= 1

− 1√2

= −√

2

Example 6

Find the value(s) of x between 0 and 2� for which

a sec x = −2 b cot x = −1

Solution

a sec x = −2∴ 1

cos x= −2

∴ cos x = −12

∴ x = � − �3

or x = � + �3

i.e., x = 2�3

or x = 4�3

b cot x = −1implies

tan x = −1∴ x = � − �

4or x = 2� − �

4

i.e. x = 3�4

or x = 7�4

Using the TI-NspireCheck that the calculator is in Radian mode.

Use solve( ) from the Algebra menu (b 31) as shown.SAM

PLE


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Using the Casio ClassPada Enter and highlight the equation

1

cos(x)= −2,

tap Interactive, Equation/inequation,solve and ensure the variable is set to x.

b Enter and highlight the equation1

tan(x)= −1,

tap Interactive, Equation/inequation,solve and ensure the variable is set to x.

The Pythagorean identityConsider a point, P(�), on the unit circle.

By Pythagoras’ theorem:

OP2 = OM2 + MP2∴ 1 = (cos �)2 + (sin �)2

Now (cos �)2 and (sin �)2 may be written as cos2 � and sin2 �.x

y

–1

–1

1

1

sinθ

cosθ MO

P(θ)

Since this is true for all values of � it is called an identity.

In particular this is called the Pythagorean identity.

cos2 � + sin2 � = 1Other forms of the identity can be derived.

Dividing both sides by cos2 � gives:

cos2 �

cos2 �+ sin

2 �

cos2 �= 1

cos2 �

1 + tan2 � = sec2 �Dividing both sides by sin2 � gives:

cos2 �

sin2 �+ sin

2 �

sin2 �= 1

sin2 �

cot2 � + 1 = cosec2 �

SAM

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Example 7

a If cosec x = 74

, find cos x . b If sec x = −32

, find sin x where�

2≤ x ≤ �.

Solution

a Since cosec x = 74, sin x = 4

7Now cos2 x + sin2 x = 1so cos2 x + 16

49= 1

∴ cos2 x = 3349

∴ cos x = ±√

33

7

b Since sec x = −32, cos x = −2

3

cos2 x + sin2 x = 1∴ 4

9+ sin2 x = 1

∴ sin x = ±√

5

3

For P(x) in the 2nd quadrant, sin x is

positive

∴ sin x =√

5

3

Example 8

If sin � = 35

and�

2< � < �, find the value of cos � and tan �.

Solution

Since cos2 � + sin2 � = 1then cos2 � + 3

2

52= 1

∴ cos2 � = 1 − 925

= 1625

∴ cos � = −45

since�

2< � < �

∴ tan � = −34

as tan � = sin �cos �

SAM

PLE


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Example 9

Prove the identity1

1 − cos � +1

1 + cos � = 2 cosec2 �

Solution

LHS = 11 − cos � +

1

1 + cos �= 1 + cos � + 1 − cos �

1 − cos2 �= 2

1 − cos2 �= 2

sin2 �

= 2 cosec2 �= RHS

Exercise 11E

1 Find the exact value of each of the following.Example 5

a cot3�

4b cot

5�

4c sec

5�

6d cosec

�

2

e sec4�

3f cosec

13�

6g cot

7�

3h sec

5�

3

2 Without using a calculator write down the exact value of each of the following.

a cot 135◦ b sec 150◦ c cosec 90◦ d cot 240◦ e cosec 225◦

f sec 330◦ g cot 315◦ h cosec 300◦ i cot 420◦

3 Find the values of x between 0 and 2� for whichExample 6

a cosec x = 2 b cot x = √3 c sec x + √2 = 0 d cosec x = sec x

4 If sec � = −178

and�

2< � < �, findExample 7

Example 8 a cos � b sin � c tan �

5 If tan � = −724

and3�

2< � < 2�, find cos � and sin �.

6 Find the value of sec � if tan � = 0.4 and � is not in the 1st quadrant.

7 If tan � = 43

and � < � <3�

2, evaluate

sin � − 2 cos �cot � − sin � .

8 If cos � = 23

and � is in the 4th quadrant, find the simplest expression in surd form for

tan � − 3 sin �cos � − 2 cot � .

SAM

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9 Prove each of the following identities for suitable values of � and �.Example 9Example 9

a (1 − cos2 �)(1 + cot2 �) = 1 b cos2 � tan2 � + sin2 � cot2 � = 1c

tan �

tan �= tan � + cot �

cot � + tan � d (sin � + cos �)2 + (sin � − cos �)2 = 2

e1 + cot2 �

cot � cosec�= sec � f sec � + tan � = cos �

1 − sin �

11.6 Addition formulas and double angle formulasAddition formulasConsider a unit circle.

Let arc length AB = v unitsarc length AC = u units

∴ arc length CB = u − v units

x

y

u – v

C

A

B

(cos u, sin u)(cos v, sin v)

–1 O

1

1

Rotate �OCB so that B is coincident with A.

The point P has coordinates

(cos (u − v), sin (u − v)).Since the triangles CBO and PAO are congruent,

CB = PAu – vP

A

O–1x

1

(1, 0)

(cos(u – v), sin(u – v))

y

Applying the coordinate distance formula

CB2 = (cos u − cos v)2 + (sin u − sin v)2= 2 − 2(cos u cos v + sin u sin v)

PA2 = (cos (u − v) − 1)2 + (sin (u − v) − 0)2= 2 − 2(cos (u − v))

Equating these

2 − 2(cos u cos v + sin u sin v) = 2 − 2(cos (u − v))∴ cos (u − v) = cos u cos v + sin u sin v

Using the TI-NspireUse tExpand( )from the Trigonometrysubmenu of the Algebra menu (b 3

1) as shown.SAM

PLE


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Replacing v with −vcos (u − (−v)) = cos u cos (−v) + sin u sin (−v)

From symmetry properties

cos (−�) = cos �sin (−�) = − sin �

cos (u + v) = cos u cos v − sin u sin v

Example 10

Evaluate cos 75◦.

Solution

cos 75◦ = cos (45◦ + 30◦)= cos 45◦ cos 30◦ − sin 45◦ sin 30◦

= 1√2

·√

3

2− 1√

2· 1

2

=√

3 − 12√

2

=√

3 − 12√

2×

√2√2

=√

6 − √24

Replacing u with�

2− u in cos (u − v)

∴ cos((�

2− u

)− v

)= cos

(�2

− u)

cos v + sin(�

2− u

)sin v

Applying symmetry properties

sin � = cos(�

2− �

)and cos � = sin

(�2

− �)

∴ cos(�

2− (u + v)

)= sin u cos v + cos u sin v

∴ sin (u + v) = sin u cos v + cos u sin vReplacing v with −v

sin (u − v) = sin u cos (−v) + cos u sin (−v)∴ sin (u − v) = sin u cos v − cos u sin v

Example 11

Evaluate

a sin 75◦ b sin 15◦.

SAM

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Solution

a sin 75◦ = sin (30◦ + 45◦)= sin 30◦ cos 45◦ + cos 30◦ sin 45◦

= 12

· 1√2

+√

3

2· 1√

2

= 1 +√

3

2√

2

= 1 +√

3

2√

2×

√2√2

=√

2 + √64

b sin 15◦ = sin (45◦ − 30◦)= sin 45◦ cos 30◦ − cos 45◦ sin 30◦

= 1√2

·√

3

2− 1√

2· 1

2

=√

3 − 12√

2

=√

3 − 12√

2×

√2√2

=√

6 − √24

Addition formula for tangent

Also tan (u + v) = sin (u + v)cos (u + v)

= sin u cos v + cos u sin vcos u cos v − sin u sin v

Divide the numerator and denominator by cos u cos v �= 0

tan (u + v) = tan u + tan v1 − tan u tan v

Similarly it can be shown that

tan (u − v) = tan u − tan v1 + tan u tan v

Example 12

If tan u = 4 and tan v = 35

and u and v are acute angles, show that u − v = �4

.

SAM

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Solution

tan (u − v) = tan u − tan v1 + tan u tan v

=4 − 3

5

1 + 4 × 35

=17

517

5

= 1

∴ u − v = �4

Note: tan � is a one-to-one function for 0 < � <�

2

Derivation of the addition formulas usingrotation about the originThe use of matrices to describe rotations about

the origin has been discussed in Section 10.11.

We can use matrices as an alternative method to

derive the addition formulas. Consider, for

example, the point with coordinates

(cos (u + v), sin (u + v)), which can be regardedas the image of a point with coordinates (cos u, sin u)

under a rotation of vc in an anticlockwise

direction around the origin.

x

y

O

v

u

(cos(u + v), sin(u + v))(cos u, sin u)

The matrix that defines a rotation of v radians anticlockwise about the origin is given by[cos v −sin vsin v cos v

]

i.e.,

[x ′

y′

]=

[cos v −sin vsin v cos v

] [x

y

]

becomes

[cos (u + v)sin (u + v)

]=

[cos v −sin vsin v cos v

] [cos u

sin u

]

i.e., cos (u + v) = cos v cos u − sin v sin u or cos (u + v) = cos u cos v − sin u sin vand sin (u + v) = sin v cos u + cos v sin u or sin (u + v) = sin u cos v + cos u sin v

Similarly, consider a point

(cos (u − v), sin (u − v)), which can beregarded as the image of a point with

coordinates (cos u, sin u) under a rotation

of vc in a clockwise direction around the

origin. O

u

v (cos(u – v), sin(u – v))(cos u, sin u)

x

ySAM

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The matrix that defines a rotation of v radians clockwise about the origin is given by[cos (−v) −sin (−v)sin (−v) cos (−v)

]=

[cos v sin v

−sin v cos v

]

i.e.,

[x ′

y′

]=

[cos v sin v

−sin v cos v

] [x

y

]

becomes

[cos (u − v)sin (u − v)

]=

[cos v sin v

−sin v cos v

] [cos u

sin u

]

i.e., cos (u − v) = cos v cos u + sin v sin u or cos (u − v) = cos u cos v + sin u sin vand sin (u − v) = −sin v cos u + cos v sin u or sin (u − v) = sin u cos v − cos u sin v

Double angle formulascos (u + v) = cos u cos v − sin u sin v

Replacing v with ucos (u + u) = cos u cos u −sin u sin ucos 2u = cos2 u −sin2 u

= 2 cos2 u − 1= 1 − 2 sin2 u

since sin2 u = 1 − cos2 usince cos2 u = 1 −sin2 u

Similarly, replacing v with u in sin (u + v) = sin u cos v + cos u sin v∴ sin 2u = sin u cos u + cos u sin u

sin 2u = 2 sin u cos u

Replacing v with u in tan (u + v) = tan u + tan v1 − tan u tan v

∴ tan (u + u) = tan u + tan u1 − tan u tan u

tan 2u = 2 tan u1 − tan2 u

Example 13

If tan � = 43

and 0 < � <�

2, evaluate

a sin 2� b tan 2�

SAM

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5 4

3θ

Solution

a sin � = 45

and cos � = 35

∴ sin 2� = 2 sin � cos �= 2 × 4

5× 3

5= 24

25

b tan 2� = 2 tan �1 − tan2 �

=2 × 4

3

1 − 169

=8

3−79

= −247

Example 14

Prove each of the following identities.

a2 sin � cos �

cos2 � −sin2 � = tan 2� bsin �

sin �+ cos �

cos �= 2 sin (� + �)

sin 2�

c1

cos � + sin � +1

cos � −sin � = tan 2� cosec �

Solution

a LHS = 2 sin � cos �cos2 � −sin2 �

= sin 2�cos 2�

= tan 2�= RHS

Note: Identity holds when cos 2� �= 0

b LHS = sin �sin �

+ cos �cos �

= sin � cos � + cos � sin �sin � cos �

= sin (� + �)1

2sin 2�

= 2 sin (� + �)sin 2�

Note: Identity holds when sin 2� �= 0

c LHS = 1cos � + sin � +

1

cos � −sin �= cos � −sin � + cos � + sin �

cos2 � −sin2 �= 2 cos �

cos 2�

But 2 cos � = 2 sin � cos �sin �

= sin 2�sin �

∴ LHS = sin 2�cos 2� sin �

= tan 2�sin �

= tan 2� cosec �Note: Identity holds when cos 2� �= 0

SAM

PLE


P1: FXS/ABE P2: FXS



Sometimes the easiest way to prove two expressions are equal is to simplify each of them. This

is demonstrated in the following example.

Example 15

Prove that (sec A − cos A)(cosecA −sin A) = 1tan A + cot A

Solution

LHS = (sec A − cos A)(cosecA −sin A) RHS = 1tan A + cot A

=(

1

cos A− cos A

) (1

sin A−sin A

)= 1

sin A

cos A+ cos A

sin A

= 1 − cos2 A

cos A× 1 −sin

2 A

sin A= 1

sin2 A + cos2 Acos A sin A

= sin2 A cos2 A

cos A sin A= cos A sin A

sin2 A + cos2 A= cos A sin A = cos A sin A

LHS = RHS

Exercise 11F

1 By making use of the appropriate addition formula find the exact values for each of theExample 10

following.

a cos 15◦ b cos 105◦

2 By making use of the appropriate addition formula find exact values for each of theExample 11

following.

a sin 165◦ b tan 75◦

3 Find exact values of

a cos5�

12b sin

11�

12c tan

−�12

4 If sin u = 1213

and sin v = 35

, evaluate sin (u + v). (Note: There is more than one answer.)Example 12

5 Simplify the following.

a sin(

� + �6

)b cos

(� − �

4

)c tan

(� + �

3

)d sin

(� − �

4

)6 Simplify

a cos (u − v) sin v + sin (u − v) cos v b sin (u + v) sin v + cos (u + v) cos v

SAM

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P1: FXS/ABE P2: FXS



7 If sin � = −35

and � is in the 3rd quadrant and cos � = −513

and � is in the 2nd quadrant,Example 13

evaluate each of the following without using a calculator.

a cos 2� b sin 2� c tan 2� d sec 2�

e sin (� + �) f cos (� − �) g cosec (� + �) h cot 2�

8 If tan u = 43

and tan v = 512

and both u and v are acute angles evaluate:

a tan (u + v) b tan 2u c cos (u − v) d sin 2u

9 If sin � = 35

and sin � = 2425

and�

2< � < � < � evaluate

a cos 2� b sin (� − �) c tan (� + �) d sin (2�)

10 If sin � = −√

3

2and cos � = 1

2evaluate

a sin 2� b cos 2�

11 Simplify each of the following expressions.

a (sin � − cos �)2 b cos4 � −sin4 �

12 Prove the following identities,Examples 14, 15

a√

2 sin(

� − �4

)= sin � − cos � b cos

(� − �

3

)+ cos

(� + �

3

)= cos �

c tan(

� + �4

)tan

(� − �

4

)= −1 d cos

(� + �

6

)+ sin

(� + �

3

)= √3 cos �

e tan(

� + �4

)= 1 + tan �

1 − tan � fsin (u + v)cos u cos v

= tan v + tan u

gtan u + tan vtan u − tan v =

sin (u + v)sin (u − v) h cos 2� + 2 sin

2 � = 1

i sin 4� = 4 sin � cos3 � − 4 cos � sin3 � j 1 −sin 2�sin � − cos � = sin � − cos �

11.7 a cos x + b sin xIn Section 11.2 the method of addition of ordinates was used in the plotting of the sums of

circular functions. In this section it will be shown how functions with rule of the form

f (x) = a cos x + b sin x may have the rule written in terms of a single circular function.First write

a cos x + b sin x = √a2 + b2(

a√a2 + b2 cos x +

b√a2 + b2 sin x

)

= √a2 + b2 (cos � cos x + sin � sin x)where cos � = a√

a2 + b2 and sin � =b√

a2 + b2Let r = √a2 + b2 and thus

a cos x + b sin x = r cos (x − �)

SAM

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P1: FXS/ABE P2: FXS



Similarly it may be shown that

a cos x + b sin x = r sin (x + �)

where r = √a2 + b2, sin � = a√a2 + b2 and cos � =

b√a2 + b2

Example 16

Express cos x − √3 sin x in the form r cos (x − �) and hence find the range of the functionwith rule f (x) = cos x − √3 sin x , and the maximum and minimum values of the function.

Solution

a = 1, b = −√

3 ∴ r = √1 + 3 = 2also cos � = a

r= 1

2and sin � = b

r= −

√3

2∴ � = −�

3∴ cos x −

√3 sin x = 2 cos

(x + �

3

)∴ Range of f is [−2, 2]

The maximum and minimum values of f are 2 and −2 respectively.

Using the TI-NspireUse tCollect( ) from the Trigonometrysubmenu of the Algebra menu (b 3

2) as shown.

Example 17

Solve cos x − √3 sin x = 1 for x ∈ [0, 2�].

Solution

From Example 16, cos x −√

3 sin x = 2 cos(

x + �3

)∴ 2 cos

(x + �

3

)= 1

cos(

x + �3

)= 1

2

x + �3

= �3

,5�

3,

7�

3

x = 0, 4�3

, 2�

SAM

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P1: FXS/ABE P2: FXS



Using the TI-NspireUse Solve( ) from the Algebra menu (b31) as shown.

The symbol ≤ can be found in thecatalog ( 4), by typing , or by

typing / .

Example 18

Express√

3 sin 2x − cos 2x in the form r sin (2x + �).

Solution

A slightly different technique is used.

Let√

3 sin 2x − cos 2x = r sin (2x + �)Then

√3 sin 2x − cos 2x = r [sin 2x cos � + cos 2x sin �]

This is to hold for all x.

If x = �4

,√

3 = r cos � ... 1

If x = 0, −1 = r sin � ... 2Squaring and adding 1 and 2 gives

r2 cos2 � + r2 sin2 � = 4i.e., r2 = 4

∴ r = ±2The positive solution is taken. Substituting in 1 and 2 gives

∴√

3

2= cos � and − 1

2= sin �

∴ � = −�6

∴√

3 sin 2x − cos 2x = 2 sin(

2x − �6

)Expand the right hand side of the equation to verify.SA

MPL

E


P1: FXS/ABE P2: FXS



Exercise 11G

1 Find the maximum and minimum values of the following.Example 16

a 4 cos x + 3 sin x b √3 cos x + sin x c cos x −sin xd cos x + sin x e 3 cos x + √3 sin x f sin x − √3 cos xg cos x − √3 sin x + 2 h 5 + 3 sin x − 2 cos x

2 Solve each of the following for x ∈ [0, 2�] or � ◦ ∈ [0, 360].Example 17a sin x − cos x = 1 b √3 sin x + cos x = 1c sin x − √3 cos x = −1 d 3 cos x − √3 sin x = 3e 4 sin �◦ + 3 cos �◦ = 5 f 2√2 sin �◦ − 2 cos �◦ = 3

3 Write√

3 cos 2x −sin 2x in the form r cos (2x + �).

4 Write cos 3x −sin 3x in the form r sin (3x − �).Example 18

5 Sketch the graphs of the following, showing one cycle.

a f (x) = sin x − cos x b f (x) = √3 sin x + cos xc f (x) = sin x + cos x d f (x) = sin x − √3 cos x

SAM

PLE


P1: FXS/ABE P2: FXS


Rev

iew


Chapter summary

Further symmetry properties: complementary angles

sin(�

2− �

)= cos �

sin(�

2+ �

)= cos �

cos(�

2− �

)= sin �

cos(�

2+ �

)= −sin �

Addition of ordinates

θ

y

3210

y1 = 2 sin x

y2 = 3 cos 2x y = 2 sin x + 3 cos 2x

π2

2ππ

–3–2–1

–4–5

23π

Graph of tangent function

y

π

1

22ππ

4

π2

3π2 2

5π –1

–πθ

y = tan θ

period = π

General solution of circular function equations

If cos (x) = a, x = 2n� ± cos−1(a), where n ∈ Z and a ∈ [−1, 1]If tan (x) = a, x = n� + tan−1(a), where n ∈ Z and a ∈ RIf sin (x) = a, x = 2n� + sin−1(a), or

x = (2n + 1)� −sin−1 (a), where n ∈ Z and a ∈ [−1, 1]Reciprocal circular functions

secant � = sec � = 1cos �

cosecant � = cosec � = 1sin �

cotangent � = cot � = 1tan �

, sin � �= 0 and cos � �= 0

SAM

PLE


P1: FXS/ABE P2: FXS


Review


Pythagorean identity

cos2 � + sin2 � = 11 + tan2 � = sec2 �cot2 � + 1 = cosec2 �

Addition formulas

cos (u − v) = cos u cos v + sin u sin vcos (u + v) = cos u cos v −sin u sin vsin (u + v) = sin u cos v + cos u sin vsin (u − v) = sin u cos v − cos u sin vtan (u + v) = tan u + tan v

1 − tan u tan vtan (u − v) = tan u − tan v

1 + tan u tan vDouble angle formulas

cos 2u = cos2 u −sin2 u= 2 cos2 u − 1= 1 − 2 sin2 u

sin 2u = 2 sin u cos utan 2u = 2 tan u

1 − tan2 ua cos x + b sin x can be written as r cos (x − �)

where r =√

a2 + b2 and cos � = a√a2 + b2 and sin � =

b√a2 + b2

It can also be written as r sin (x + �)

where r =√

a2 + b2 and sin � = a√a2 + b2 and cos � =

b√a2 + b2

Multiple-choice questions

1 cosec x −sin x is equal toA cos x cot x B cosec x tan x C 1 −sin2 xD sin x cosec x E

1 −sin xsin x

2 If cos x = −13

, the possible values of sin x are

A−2√2

3,

2√

2

3B

−23

,2

3C

−89

,8

9

D−√2

3,

√2

3E

1

2,−12

SAM

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P1: FXS/ABE P2: FXS


Rev

iew


3 If cos � = ab

and 0 < � <�

2, then tan � in terms of a and b is

A

√a2 + b2

bB

√b2 − a2

aC

a√b2 − a2

Da√

b2 + a2 Ea

b√

b2 + a24 The magnitude of ∠ABX is �, AX = 4 cm, XC = x cm

and BC = 2 cm. In terms of x, tan � is equal toA

8

(x + 2)2 B4

xC 8 − x D 8 + xE

8√x2 + 4

2 cm

θ

4 cm

x cm

X

C

A

B

5 For�

2< A < � and � < B <

3�

2, with cos A = t and sin B = t , sin (B + A) is equal to

A 0 B 1 C 2t2 − 1 D 1 − 2t2 E −1

6sin 2A

cos 2A − 1 is equal toA cot 2A − 1 B sin 2A + sec 2A C sin A

cos A − 1D sin 2A − tan 2A E −cot A

7 sin(�

2− x

)is not equal to

A cos (2� − x) B −sin(

3�

2+ x

)C sin x D cos (−x) E sin

(�2

+ x)

8 (1 + cot x)2 + (1 − cot x)2 is equal toA 2 + cot x + 2 cot 2x B 2 C −4 cot x D 2 + cot 2x E 2cosec2x

9 If sin 2A = m and cos A = n, tan A in terms of m and n is equal toA

m

2n2B

n

mC

2n

m2D

2n

mE

2n2

m10 −cos x + sin x , in the form r sin (x + �) where r > 0, is

A√

2 sin(

x + �4

)B −sin

(x + �

4

)C

√2 sin

(x + 5�

4

)

D√

2 sin

(x + 7�

4

)E

√2 sin

(x + 3�

4

)

Short-answer questions (technology-free)

1 Prove each of the following identities.

a sec � + cosec � cot � = sec � cosec2 � b sec � −sin � = tan2 � + cos2 �

sec � + sin �2 Find the maximum and minimum values of each of the following.

a 3 + 2 sin � b 1 − 3 cos � c 4 sin 32

�

d 2 sin21

2� e

1

2 + cos �

SAM

PLE


P1: FXS/ABE P2: FXS


Review


3 Find the values of �, � ∈ [0, 2�], for whicha sin2 � = 1

4b sin 2� = 1

2c cos 3� =

√3

2

d sin2 2� = 1 e tan2 � = 13

f tan 2� = −1g sin 3� = −1 h sec 2� = √2

4 Solve the equation tan � = 2 sin � for values of � from 0◦ to 360◦.5 If sin A = 5

13, sin B = 8

17where A and B are acute, find

a cos (A + B) b sin (A − B) c tan (A + B)6 Find

a cos 80◦ cos 20◦ + sin 80◦ sin 20◦ b tan 15◦ + tan 30◦

1 − tan 15◦ tan 30◦7 If A + B = �

2, find the value of

a sin A cos B + cos A sin B b cos A cos B −sin A sin B8 Find the maximum and minimum values of the function with rule

a 3 + 2 sin � b 4 − 5 cos �9 Prove each of the following.

a sin2 A cos2 B − cos2 A sin2 B = sin2 A −sin2 B b sin �1 + cos � +

1 + cos �sin �

= 2sin �

csin � − 2 sin3 �2 cos3 � − cos � = tan �

10 Given that sin A =√

5

3and that A is obtuse, find the value of each of the following:

a cos 2A b sin 2A c sin 4A

11 Prove

a1 − tan2 A1 + tan2 A = cos 2A b

sin A

1 + cos A +1 + cos A

sin A= 2

sin A

12 a Find tan 15◦ in simplest surd form.b Using the identities for sin (u ± v), express 2 sin x cos y as the sum of two sines.

13 Given f : [0, 2�] → R, f (x) = 2√3 cos x − 2 sin x , find the coordinates ofa the y intercept b the x intercepts

c the maximum point d the minimum point.

Hence sketch the graph of f (x) = 2√3 cos x − 2 sin x14 Solve for x, 0 ≤ x ≤ 2�.

a sin x + cos x = 1 b sin 12

x cos1

2x = −1

4c 3 tan 2x = 2 tan x d sin2 x = cos2 x + 1e sin 3x cos x − cos 3x sin x =

√3

2f 2 cos

(2x − �

3

)= −√3

SAM

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iew


15 Sketch graphs of

a y = 2 cos2 x b y = 1 − 2 sin(�

2− x

2

)c f (x) = tan 2x

16 It is given that tan A = 2. Find the exact value of tan �, given that tan (� + A) = 4.17 a Express 2 cos � + 9 sin � in the form r cos (� − �), where r > 0 and 0 < � < �

2b i Give the maximum value of 2 cos � + 9 sin �

ii Give the cosine of � for which this maximum occurs.

iii Find the smallest positive solution of the equation 2 cos � + 9 sin � = 1

Extended-response questions

1 The diagram shows a rectangle ABCD inside a semicircle, centre O and radius 5 cm.

∠BOA = ∠COD = �◦a Show that the perimeter, P cm, of the rectangle

is given by

P = 20 cos � + 10 sin �b Express P in the form r cos (� − �) and hence

find the value of � for which P = 16. θ° θ°A

B C

DO

5 cm 5 cm

c Find the value of k for which the area of the rectangle is k sin 2� cm2.

d Find the value of � for which the area is a maximum.

2 The diagram shows a vertical section through a tent

in which AB = 1 m, BC = 2 m and∠BAD = ∠BCD = �. CD is horizontal.

The diagram is symmetrical about the vertical AD.

a Obtain an expression for AD in terms of �.

b Express AD in the form

r cos (� − �), where r is positive.c State the maximum length of AD

and the corresponding value of �.

A

B

C D

1 m

2 m

θ

θ

d Given that AD = 2.15 m, find the value of � for which � > �.3 a Prove the identity cos 2� = 1 − tan

2 �

1 + tan2 �b i Use the result of a to show 1 + x2 = √2x2 − √2 where x = tan

(67

1

2

)◦ii Hence find the values of integers a and b such that tan

(67

1

2

)◦= a + b√2

c Find the value of tan

(7

1

2

)◦.

SAM

PLE


P1: FXS/ABE P2: FXS


Review


4 In the diagram triangle ABC has a right angle at B.

Length of BC = 1 unit.a Find in terms of �

i h1 ii h2 iii h3 iv hn

A

h1

h2h3

B C

θ

b Show that the infinite sum h1 + h2 + h3 + . . . = cos �1 −sin �

c If the infinite sum = √2, find �.5 ABCD is a regular pentagon with side length one unit.

The exterior angles of a regular pentagon each have

magnitude2�

5.

a i Show that the magnitude of ∠BCA is �5

ii Find the length of CA

b i Show the magnitude of ∠DCP is 2�5

ii Use the fact that AC = 2CQ = 2CP + PRto show that 2 cos

�

5= 2 cos 2�

5+ 1

B

CP Q R

A

D E

2π5

iii Use the identity cos 2� = 2 cos2 � − 1 to form a quadratic equation in terms ofcos

�

5iv Find the exact value of cos

�

56 a Prove each of the identities

i cos � =1 − tan2 �

2

1 + tan2 �2

ii sin � =2 tan

�

2

1 + tan2 �2

b Use the result of a to find the value of tan�

2, given 8 cos � − sin � = 4

SAM

PLE


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