chapter 3_trigonometry and circular functions

LFSE012 Sci/Eng Mathematics A Trig & Circ Fn

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CHAPTER 3 TRIGONOMETRY & CIRCULAR FUNCTIONS

3.1 INTRODUCTION

The word “Trigonometry” comes from Greek words which mean “the measurement of

triangles”. The trigonometric ratios (sine, cosine and tangent) are defined for a right-angled

triangle.

The ratios are written as sin, cos and tan.

oppositesin θ =

hypotenuse

adjacentcos θ =

hypotenuse

oppositetan θ =

adjacent

hypotenuse

opposite

θ

adjacent

3.1.1 MEASURING ANGLES

When the line OA is rotated about O, the amount of turn is called an angle.

Angles can be measured in degrees or radians.

Degree measure

When the line OA is rotated through a full revolution, it

has been rotated through 360 degrees (written 360º)

A degree is 1

of a full revolution360

th

y

A’

O A x

Angle A'OA

Radian measure

Radian measure measures angles in terms of the radius of

a circle.

Definition : When an angle cuts off an arc on a circle

equal to the radius of the circle, the amount of turn is 1

radian.

1 radian ≈ 57.3º

Notes :

1 radian is written as 1c .

If there is neither degree nor radian symbol,

assume that the angle is in radians.

y

R

1 rad

R x

The circumference of a circle is C = 2 π r. So for 1 full revolution, we turn through 2π radians.

So 2π c = 360º ; which means :

π = 180c

This can be used to convert any angle from degrees to radians, and vice versa.


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Example 1

Convert to radians : (i) 135º (ii) 60º

Solution:

π 3π π π(i) 135 135 (ii) 60 60

180 4 180 3

c c

Example 2

Convert to degrees : 5π

i ii 2.56

c

Solution:

5π 5π 180 180 450

i = 150 ii 2.5 2.5 143 2 to 4 significant figures)6 6 π π π

. (c c

c

3.1.2 EXACT VALUES FOR TRIGONOMETRIC RATIOS

30º 45º

60º 60º 45º

1 1 1

2 3 2 2 1

θ sin θ cos θ tan θ

0 0 1 0

π

6

1

2 3

2

1

3

π

4

1

2

1

2

1

π

3 3

2

1

2

3

π

2

1

0

(undefined)

Exercise 3.1

* When writing approximate answers, write to 4 significant figures

1. Express the following angles in radians:

(a) 45º (b) 270º (c) 30º (d) 450º (e) 2 revolutions

(f) 33º (g) 317º (h) 12.5º (i) 12º 15’

2. Express the following angles in degrees:

2π 7π 5π 5π 7πa b c d e

3 6 2 12 4

f 2 g 1.5 h 4.21 i 0 006 .

c c

c c c c


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3. Write the value of each of the following in exact form:

π π π πa tan b cos c sin d tan e cos 0

3 6 2 2

f sin 45 g cos 60 h tan 90

c c

4. In each of the following find the ratio in exact form (in each case, θ is an acute angle):

3a Find sin θ, given tan θ =

4

5b Find cos θ, given tan θ =

12

4c Find tan θ, given sin θ =

5

1d Find cos θ, given tan θ =

2

3.2 TRIGONOMETRIC FUNCTIONS OF ANGLES

The trigonometric ratios can also be defined in terms of a unit circle (radius 1 unit). These are

called trigonometric functions

A ray is drawn from the origin at an angle θ

from the positive x-axis. It meets the unit circle

at P.

The triangle OPQ has a hypotenuse of 1 unit.

A tangent line is drawn to the circle at (1,0)

sin θ =

cos θ =

sin θtan θ = =

cos θ

y

x

d

y

(0,1)

P(θ)

d

y

(–1,0) x (1,0) x

(0, –1)

θ

1

3.2.1 THE FOUR QUADRANTS

The unit circle can be divided into four quadrants as

shown at right.

1st quadrant : 0 < θ <

π

2

2nd

quadrant : π

2 < θ < π

3rd

quadrant : π < θ < 3π

2

4th

quadrant : 3π

2 < θ < 2π

y

2nd

1st

π – θ θ

π + θ 2π – θ x

3rd

4th


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The trigonometric functions can be defined in terms of each of these quadrants by using

symmetry.

2nd

Quadrant

sin π θ = sin θ

cos π θ = cos θ

sinθ and since tanθ =

cosθ

tan π θ = tan θ

y

P(π – θ) P(θ)

y y

θ

–x x x

π – θ

3rd Quadrant

sin π θ = sin θ

cos π θ = cos θ


cosθ

tan π θ = tan θ

y

P(θ)

y

– y

P(π + θ)

θ

–x x x

π + θ

4

th Quadrant

sin 2π θ = sin θ

cos 2π θ = cos θ


cosθ

tan 2π θ = tan θ

y

P(θ)

y

– y

P(π + θ)

θ

x x

2π – θ

Summary :

1st Quadrant : All ratios positive

2nd

Quadrant : Sine positive

3rd

Quadrant : Tangent positive

4th

Quadrant : Cosine positive

S A

T C


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3.2.2 RECIPROCAL RATIOS

Each of the trigonometric ratios has a reciprocal ratio. These ratios are called the secant,

cosecant and cotangent.

Definition

1 1 1cosecθ = secθ = cotθ =

sinθ cosθ tanθ

Example 1

Without using a calculator, find

5π 11π 3π 3π 4π

a tan b cos c sin d cosec e cot 0 f sec 3 6 2 4 3

c c

Solution:

5π π πa tan = tan 2π = tan 3

3 3 3

11π π π 3b cos = cos 2π cos

6 6 6 2

3π π πc sin = sin π = sin = 1

2 2 2

3π 1 1 1 1d cosec 2

3π π 1π4sin sinsin π 24 44

1e cot 0 = S

tan 0

c

c

.

1ince tan 0 = 0, is undefined.

tan 0

4π 1 1 1 1f sec 2

4π π 1π3cos coscos π+ 2

3 33

Exercise 3.2

1. Without using a calculator, find the value of

5π 11π 4πa cos b sin c cosec

6 6 3

7π 5πd cot e sec 0 f tan

4 3

c c

2. Without using a calculator, find the value of

a cos 225 b sin 390 c cosec 315

d cot 120 e sec 270 f tan 225


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3.3 GRAPHING TRIGONOMETRIC FUNCTIONS

The graphs of the trigonometric functions are periodic i.e they repeat at regular intervals. Each

repeating shape is called a cycle. These functions are also called circular functions.

The maximum distance from the graph to the x-axis is called the amplitude.

One cycle of the trigonometric (circular) functions sine, cosine and tangent are shown below, as

well as their key features.

y = sin x

Period : 2π (360º)

Amplitude : 1 unit

y

1

0 π

2 π 3π

2 2π x

–1

y = cos x

Period : 2π (360º)

Amplitude : 1 unit

y

1

0 π

2 π 3π

2 2π x

–1

y = tan x

Period : π (180º)

Amplitude: undefined.

Asymptotes : 3

2 2

x x

;

y

0 π2 π

3π2 2π x

x = π2 x =

3π2

The basic trigonometric graphs can be transformed in a number of ways: they can be translated,

stretched and compressed (parallel to x-, y- axes).


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3.3.1 TRANSFORMATIONS OF TRIGONOMETRIC FUNCTIONS

We will consider transformations to the sine function only. These same translations apply to

both cosine and tangent.

In function y = sin x :

replace y with y –a

Function becomes y = a + sin x

Graph is translated a units up.

y

0 π 2π x

y = sin x

y = a + sin x a


replace x with x – c

Function becomes y = sin(x – c)

Graph is translated c units to the

right.

y

0 π 2π x

y = sin x

y = sin (x – c)

c


multiply sin x by b

Function becomes y = b sin x

Graph is dilated by a factor of b

parallel to the y-axis

If b > 1, graph is stretched

If b <1, graph is compressed

If b < 0, the graph reflects about

the x-axis

y

0 π 2π x

y = sin x

y = k sin x


replace x with kx

Function becomes y = sin kx

Graph is dilated by a factor of k

parallel to the x-axis

If k > 1, graph is compressed

If k < 1, graph is stretched

y

0 π

2 π 3π

2 2π x

y = sin x

y = sin 2x

y

0 π 2π x

y = sin x

y = sin 1

2x


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Exercise 3.3

1. Plot graphs for the following functions on the same set of co-ordinate axes in the

interval 0 ≤ x ≤ 360

(a) y = cos x (b) y = 1 + cos x

(Hint: set up a table of values for x and y)

2. Plot graphs for the following functions on the same set of co-ordinate axes in the

interval 0 ≤ x ≤ 360

(a) y = sin x (b) y = sin 2x

At what values of x do the graphs intersect?

3. Sketch graphs for the following functions in the interval 0 ≤ x ≤ 360

(a) y = sin x (b) y = 2 sin x

At what values of x do the graphs intersect?

4. Sketch the graphs of y = sin x and y = cos x in the interval 0 ≤ x ≤ 360.

Use the graph to define at which quadrant the value of x is when:

(a) sin x = cos x (b) sin x = 0.5 (c) cos x = –0.5

5. (i) Sketch the graph of 3 2siny x in the interval 0≤ x ≤ 360.

(ii) Find the maximum and minimum values of y.

(iii) Use the graph to solve the equation 3 + 2 sin x = 1

6. (i) Sketch the graph of cos2 1y x in the interval 0≤ x ≤ 360.

(ii) Use the graph to solve the equation cos2 1 0x

(iii) Use the graph to solve the equation cos2 1 1x

7. (i) Sketch the graph of 2 3sin 2y x in the interval 0≤ x ≤ 360.

(ii) Find the maximum and minimum values of y.

(iii) Use the graph to solve the equation 3 + 2 sin x = 5

(iv) At which quadrant the value of x is when 3 + 2 sin x = 1


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3.4 TRIGONOMETRIC IDENTITIES

Relationships between trigonometric ratios which are true for any angle(s) are called identities.

All the identities below can be proved by use of the unit circle and Pythagoras’ Theorem.

Pythagorean Identities 2 2

2 2

2 2

sin A + cos A = 1

1 + tan A = sec A

1 + cot A = cosec A

Compound Angles

sin A + B = sinA cosB + cosA sinB

sin A B = sinA cosB cosA sinB

cos A + B = cosA cosB sinA sinB

cos A B = cosA cosB sinA sinB

tanA + tanBtan A + B =

1 tanAtanB

tanA tanBtan A B =

1 tanAtanB

Double Angles

2 2 2 2

2

sin 2A = 2 sinA cosA

cos 2A = 2 cos A 1 = 1 2 sin A = cos A sin A

2 tanAtan 2A =

1 tan A

Products as Sums and Differences

2 sinA cos B = sin A+B sin A B

2 cosA sin B = sin A+B sin A B

2 cosA cos B = cos A+B cos A B

2 sinA sin B = cos A B cos A B

Sums and Differences as Products

1 1sinA + sinB = 2 sin A+B cos A B

2 2

1 1sinA sinB = 2 cos A+B sin A B

2 2

1 1cosA + cosB = 2 cos A+B cos A B

2 2

1 1cosA cos B = 2 sin A+B sin A B

2 2


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You do not need to memorise the above identities. However you need to use the appropriate

identity from the list above to solve problems, and prove other identities.

Example 1

Find, without the use of a calculator, the value of sin A, given that cos 2A = 5

13

Solution:

2

2

2

cos 2A = 1 2 sin A

2 sin A 1 cos 2A

1sin A 1 cos 2A

2

1 5 4 1 =

2 13 13

4 2 13 sinA =

13 13

Example 2

Prove the identity: sin A+B cosB cos A+B sinB = sinA

Solution:

2 2

LHS = sin A+B cosB cos A+B sinB

= sinA cosB+ cosA sinB cosB cosA cosB sinA sinB sinB

= sinA cos B+ cosA sinB cosB cosA sinB cosB sinA sin B

= sinA

2 2

2 2

cos B sinA sin B

= sinA cos B sin B

= sinA 1

= sinA = RHS

The identity is proven

Example 3

Prove the identity: sin2A

cotA1 cos2A

Solution:

2

2

sin2A 2 sinA cosALHS = =

1 cos2A 1 1 2 sin A

2 sinA cosA =

2 sin A

cosA =

sinA

cotA = RHS

The identity is proven


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Example 4

Without using a calculator, find an exact value for : (i) cos 75º (ii) tan 105º

Solution:

o o

i cos75 cos 45 30

1 3 1 1 = cos45 cos30 sin45 sin30 =

2 22 2

2 3 13 1 =

42 2

ii tan105 tan 60 45

tan60 tan45 3 1 =

1 tan60 tan45 1 3

3 1 1 3 =

1 1 3 1 3

3 1 3 1 1 3 3 3 1 3 = =

1 3 3 31 1 3 3 1 3

2 3 2 = = 3 2

2

Exercise 3.4

For problems 1 to 7 use the following identities:

sin A + B = sinA cosB + cosA sinB sin A B = sinA cosB cosA sinB

cos A + B = cosA cosB sinA sinB cos A B = cosA cosB sinA sinB

tanA + tanBtan A + B =

1 tanAtanB

tanA tanB tan A B =

1 tanAtanB

1. Without using a calculator, find the exact value for sin 105º by using the ratios of 45º

and 60º

2. Simplify

(i) sin (A + B) + sin (A – B)

(ii) cos (A + B) – cos (A – B)

3. Prove the identity : sin (A + B)

tan A + tan B = cos A cos B

4. Prove the identity : sin A + B sin A B = sinA sinB sinA sinB

5. Prove the identity : sin A B cosB + cos A B sinB = sinA

6. Prove that π π

sin + A sin A = 2 sinA4 4

7. 4 12

If A and B are acute angles such that tan A and cos B = , find cos A B3 13

without using a calculator .


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2 2 2 2

2

sin 2A = 2 sinA cosA

cos 2A = 2 cos A 1 = 1 2 sin A = cos A sin A

2 tanAtan 2A =

1 tan A

8. Find the required expression without using a calculator :

(a) sin 2A , given 3

sinA = 5

(b) cos 2A , given 3

cosA = 5

(c) tan 2A , given 1

tanA = 2

9. Prove the identity : 1 + cos2A + cosA

= cotAsin2A + sinA

10. Prove the identity : sin2A

= tanA1 + cos2A

11. Prove the identity : 21 + cosA A = cot

1 cosA 2

12. Prove the identity : sin2A

= cotA1 cos2A

13. Prove the identity : sin2A cos2A

= secAsinA cosA

14. Prove the identity : 1 + sinA cosA A

= tan1 + sinA cosA 2


1 1sinA + sinB = 2 sin A+B cos A B

2 2

1 1sinA sinB = 2 cos A+B sin A B

2 2

1 1cosA + cosB = 2 cos A+B cos A B

2 2

1 1cosA cos B = 2 sin A+B sin A B

2 2

15. Prove the identity : 2 sin 4A cosA = sin 5A + sin 3A

16. Prove the identity : 2 cos 5A cos 2A = cos 3A + cos 7A


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17. Prove the identity : cos 2A cos 4A

= tan 3Asin 4A sin 2A

18. Prove the identity : cos A cos 3A

= tan Asin 3A sin A

3.5 TRIGONOMETRIC EQUATIONS

An equation containing a trigonometric term and an “equals” sign is called a trigonometric

equation. These equations may or may not have solutions.

The diagram below shows the graph of y = sin x for –540 ≤ x ≤ 540

The lines y = 1.2; y = 0.5; and y = –1 have also been drawn.

-450 -3 60 -27 0 -180 -90 90 1 80 27 0 360 4 50

-1 .5

-1

-0 .5

0 .5

1

1 .5

x

y

y = 1.2

y = sin xº

y = 0.5

y = –1

For the interval –540≤ x ≤540 :

There is no intersection between y = sin xº and y = 1.2

sin xº = 1.2 has no solution.

There are 6 intersection points between y = sin xº and y = 0.5

sin xº = 0.5 has 6 solutions in this domain

There are 3 intersection points between y = sin xº and y = –1

sin xº = –1 has 3 solutions in this domain


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We will look at four types of trigonometric equations.

TYPE 1 Simple linear equations

Example 1

Find the solution of the equation cos x = – 0.3692 for 0º ≤ x ≤ 360º

Solution:

By calculator cos x = 0.3692 x = 68.3º

Cosine is negative in the second and third quadrants

x = 180º – 68.3º or x = 180º + 68.3º

x = 111.7º or 248.3º

Example 2

Find the solution of the equation 3 tan (2x + 10º) + 5.395 = 0 for 0º ≤ x ≤ 360º

Solution:

NOTE : 0º ≤ x ≤ 360º 10º ≤ 2x+10 ≤ 730º

3 tan (2x + 10º) + 5.395 = 0

3 tan (2x + 10º) = – 5.395

tan (2x + 10º) = 5 395

1 79833

..

tan is negative in 2

nd and 4

th quadrants.

2x + 10º = (180º – 60.92º) , (360º – 60.92º), (540º – 60.92º) , (720º – 60.92º)

2x + 10º = 119.08º , 299.08º , 479.08º , 659.08º

2x = (119.08º – 10º) , (299.08º – 10º), (479.08º– 10º) , (659.08º – 10º)

2x = 109.08º , 289.08º , 469.08º , 649.08º

x = 54.5º , 144.5º , 234.5 º , 324.5º.

Exercise 3.5A

1. Find the solutions of the equation 4 sin = 3x , for 0º ≤ x ≤ 360º

2. Solve the equation sin 4 + 20 = 0x , for 0º ≤ x ≤ 360º

3. Solve the equation π

2 cos 2 = 13

x

for 0º ≤ x ≤ 360º

4. Solve, for 0 ≤ θ ≤ 2π , θ : 3 tan 2θ = 1


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TYPE 2 Equations reducible to quadratic form

The quadratic equation is factorised into two linear equations.

Example 1

Find the solution of the equation 3 cos 2θ + cos θ + 1 = 0 for 0 ≤ θ ≤ 2π

Solution:

3 cos 2θ + cos θ + 1 = 0

2

2

2

3 2 cos θ 1 + cos θ + 1 = 0

6 cos θ 3 + cos θ + 1 = 0

6 cos θ + cos θ 2 = 0

3 cosθ + 2 2 cosθ 1 = 0 **we have 2 linear equations to solve

If 3 cosθ + 2 = 0

2 cosθ =

3

nd rd **cos is negative in 2 and 3 quadrants

θ = π 0.8411 or π 0.8411 ** angle is in radians

= 2.301 or 3.983

If 2 cosθ 1 = 0

1cosθ =

c c

st th **cos is positive in 1 and 4 quadrants2

π π θ = or 2π

3 3

π 5π = or

3 3

π 5πθ = , 2.301 , 3.983 ,

3 3

c c

Exercise 3.5B

1. Find, for 0 ≤ θ ≤ 2π , 2θ : 2 sin θ + cos θ = 1

2. Find the solution of the equation sin x – cos 2x = 0 for 0º ≤ x ≤ 360º

3. Find the solution of the equation sin 2θ = tan θ for 0 ≤ θ ≤ 2π

4. Find, for 0 ≤ θ ≤ 2π , 3θ : tan θ = tan θ

*5. Find the solution of the equation 6 tan2

θ – 4 sin2 θ = 1 for 0 ≤ θ ≤ 2π


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TYPE 3 sin A ± sin B =0 or cos A ± cos B =0

The sum or difference is changed into a product.

Example 1

Solve for x the equation sin 3 = sin for the domain 0 360x x x

Solution:

sin 3 = sin

sin 3 sin = 0

1 12 cos 3 sin 3 0

2 2

2 cos 2 sin = 0

If cos 2 = 0

2 = 90 , 360 90 , 360 90 , 720 90

2 = 90 ,

x x

x x

x x x x

x x

x

x

x

( ) ( )

270 , 450 , 630

= 45 , 135 , 225 , 315

If sin = 0

= 0 , 180 , 360

Solution is = 0 , 45 , 135 , 180 , 225 , 315 , 360

x

x

x

x .

Exercise 3.5C

1. Find the solution of the equation cos 3θ cos θ = 0 for 0 ≤ θ ≤ 2π

2. Find the solution of the equation 3

sin sin = sin 2 2

x xx , for 0º ≤ x ≤ 360º

3. Find , for 0º ≤ x ≤ 360º , θ : sin 5 = sin 3 x x

4. Solve the equation cos 4 cos 2 = cos 3x x x , for 0º ≤ x ≤ 360º


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TYPE 4 a sin θ ± b cos θ = C

a sin θ ± b cos θ is changed into the form R sin (θ ± α ) or R cos (θ ± α )

Example 1

Solve for x the equation 3 sin + 4 cos = 3 for 0 360x x x

Solution:

3 sin + 4 cos = 3

Let R sin + α = 3 sin + 4 cos

R sin cos α + R cos sin α = 3 sin + 4 cos x

Equate the coefficients of sin , cos :

R cos α = 3 and

x x

x x x

x x x

x x

2 2 2 2

2 2 2 2

R sin α = 4

Square both equations :

R cos α = 9 and R sin α = 16

Adding the equations :

R cos α + R sin α = 9 + 16

2 2 2

2

R cos α + sin α = 25

R = 25

R = 5

5 cos α = 3 and 5 sin α = 4

3 4 cos α = and sin α =

5 5

α = 53.13

3 sin + 4 cos =x x

st nd

5 sin + 53.13

5 sin + 53.13 = 3

3 sin + 53.13 = 0 6 ** sin positive in 1 and 2 quadrants

5

+ 53.13 = 36.87 , 180 36.87 , 360 + 36.87

x

x

x

x

.

+ 53.13 = 36.87 , 143.13 , 396.87

= 36.87 53.13 , 143.13 53.13 , 396.87 53.13

= 16 3 , 90 , 343.7

Solution is

x

x

x

.

= 90 , 343.7x

Exercise 3.5D

1. Find the solution of the equation 4 sin 3 cos = 2x x , for 0 ≤ x ≤ 360º

2. Find the solution of the equation 7 cos 24 sin = 16x x , for 0 ≤ x ≤ 360º

3. Solve the equation sin θ cos θ = 1 , for 0 ≤ θ ≤ 2π


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3.6 INVERSE TRIGONOMETRIC FUNCTIONS

Consider the statement: “the square of 3 is 9”.

The reverse statement would be: “the square root of 9 is 3”

The operations “square” and “square root” are inverse operations.

We are now going to consider inverse trigonometric functions.

Principal values of trigonometric functions

The inverse of a function can exist only if the function is a 1:1 function.

We can make sine, cosine and tangent functions into 1:1 functions by restricting the domain of

each function. They are called the Principal Value functions.

y = Sin x

Domain π π

2 2

x

y

1

y = Sin x

π

2

π

2

x

–1

y = Cos x

Domain 0 πx

y

1

y = Cos x

π x

–1

0

y = Tan x

Domain π π

2 2

x

The values π π

2 2

, are not included.

The lines x = ±π

2 are asymptotes

y

y = Tan x

π

2

π

2

x


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3.6.1 The Inverse Sine Function

If x = Sin y then 1 = Siny x is the inverse sine function for –1 ≤ x ≤ 1

y

1

y = Sin x

π

2

π

2

x

–1

y π

2

y = Sin–1

x

–1 1 x

π

2

Note: 1Sin can be written as Arsin or Arsine or Arcsine x x x x

**** 1 1 = Sin does not mean =

Sin y x y

x

******

The uppercase S indicates the principal value of the inverse function.

For example: 1 1 πSin =

2 6

. This is the principal value.

However 1 1sin

2

has many solutions: 1 1 11π 7π π 5π 13π 17πsin = ....., , , , ,

2 6 6 6 6 6 6, , ....

Example 1

Use calculators to find:

(i) 1Sin 0.98 (in degrees) (ii) 1Sin 0.47 (in radians)

(iii) 1Sin 1 01. (in degrees)

Solution:

(i) 1Sin 0.98 = 78.52º (to 4 sig.fig.)

(ii) 1Sin 0.47 = –0.4893 (to 4 sig.fig.)

(iii) 1Sin 1 01. does not exist (error message)

Example 2

Find, without calculators, the value of:

(i) 1 3Sin

2

(ii) 1 1Sin

2

(iii) Arsin 1.2 (iv) 1 3πSin sin

4

(v) 1Sin sin π

Solution:

(i) 1 3 π π πSin value must be between and

2 3 2 2

(ii) 1 1 π π πSin value must be between and

2 6 2 2

(iii) Arsin 1.2 does not exist. The value of must lie in the interval 1 1 x ,

(iv) 1 13π 1 πSin sin = Sin =

4 42

(v) 1 1Sin sin π = Sin 0 = 0 not π !!


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3.6.2 The Inverse Cosine Function

If x = Cos y then 1 = Cosy x is the inverse cosine function for –1 ≤ x ≤ 1

y

1

y = Cos x

π x

–1

0

y

π

y = Cos–1

x

–1 0 1 x

Note: 1Cos can be written as Arcos x x

The uppercase C indicates the principal value of the inverse function.

For example: 1 1 πCos =

2 3


However 1

arcos2

means every angle which has a cosine value of 1

2.

i.e. 1 5π π π 5π 7π

arcos = ..... , , , , , ,........2 3 3 3 3 3

Example 1


(i) 1Cos 0.77 (in degrees) (ii) 1Cos 1 (in radians)

(iii) 1Cos 0.53 (in radians) (iv) 1sin Cos 0 02. (in degrees)

Solution:

(i) 1Cos 0.77 = 39.65 (to 4 sig.fig.)

(ii) 1Cos 1 3.142c (to 4 sig.fig.)

(iii) 1Cos 0.53 = 2.129c (to 4 sig.fig.)

(iv) 1sin Cos 0 02 = sin 90.11 = 0.9998. (to 4 sig.fig.)

Example 2

Find, without the aid of calculators, the value of:

(i) 1 1Cos

2

(ii) 1

sin Arcos 2

(iii) 1 1cos Sin

3


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Solution:

(i) 1 1 1

Let = Cos cos = and 0 π2 2

3π π = not

4 4

y y y

y

(ii)

1 1 πLet = Arcos cos = =

2 2 3

1 π Arcos =

2 3

1 π 3sin Arcos = sin =

2 3 2

y y y

(iii)

1 1 1 π πLet Sin = sin = and

3 3 2 2

Use Pythagoras' Theorem (see triangle)

2 2cos =

3

x x x

x

3

1

x

2 2

3.6.3 The Inverse Tangent Function

If x = Tan y then 1 = Tany x is the inverse tangent function for x R

y

y = Tan x

π

2

π

2

x

y π

2

y = Tan–1

x

x

π

2

Note: 1Tan can be written as Artan x x

The uppercase T indicates the principal value of the inverse function.

For example: 1 πTan 1 =

4


However artan 1 means every angle which has a tan value of 1.

i.e. 7π 3π π 5π 9π

artan 1 = ..... , , , , , ,........4 4 4 4 4


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Example 1


(i) 1Tan 2.56 (in degrees) (ii) 1Tan 0 987. (in radians)

(iii) 1sin Tan 0.53 (in radians) (iv) 1tan Tan 0 002. (in degrees)

Solution:

(i) 1Tan 2.56 = 68.66 (to 4 sig.fig.)

(ii) 1Tan 0 987 = 0 7789. .c (to 4 sig.fig.)

(iii) 1Sin Tan 0.53 = Sin 0 48735 = 0 4683. .c (to 4 sig.fig.)

(iv) 1tan Tan 0 002 = tan 0 11459 = 0 002. . .

Example 2

Without calculators, evaluate:

(i) 1tan Tan 1 (ii) 1 2πTan tan

3

Solution:

(i) 1 πtan Tan 1 = tan = 1

4

(ii) 1 12π πTan tan = Tan 3 =

3 3

Exercise 3.6

1. Without using calculator, find (where they exist):

(i) 1Sin 1 (ii) 1Sin 0 (iii) 1 1Cos

2

(iv) 1 3Sin

2

(v) 1Cos 2 (vi) 1Sin sin 0 5. (vii) π

Arcos cos 6

2. Evaluate without the use of calculators:

(i) 1 1cos Sin

2

(ii) 1Tan tan 225 (iii) 1cos Tan 3

3. Without calculators, evaluate, using trigonometric identities if needed:

(i) 1 13 3sin Sin + sin Sin

5 5

(ii) 1 3

sin 2 Tan4

4. Show, without using a calculator, that 1 1 3 πTan 4 Tan =

5 4


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ANSWERS

Exercise 3.1

1. (a) π

4 (b)

3π

2 (c)

π

6 (d)

5π

2 (e) 4π

(f) 0.5760c (g) 5.533

c (h) 0.2182

c (i) 0.2138

c

2

a 120 b 210 c 450 d 75 e 315

f 114 6 g 85 94 h 241 2 i 0 3438

. . . .

3. 3 1 1

a 3 b c 1 d undefined e 1 f g h undefined 2 22

4. 3 12 4 2 5

a b c d 5 13 3 5

Exercise 3.2

1. 3 1 2

a b c d 1 e 1 f 32 2 3

2. 1 1 1

a b c 2 d e undefined f 122 3

Exercise 3.3

1.

y

y = 1 + cos x

1

y = cos x

0 90 180 270 360 x

–1

–2

2.

y

1 y = sin x y = sin 2x

0 90 180 270 360 x

–1

Graphs intersect at 0º, 60º,180º, 300,

360

3.

y

y = 2 sin x

1

y = sin x

0 90 180 270 360 x

–1

–2

Graphs intersect at 0º, 180º,360º

4.

y

1 y = sin x y = cos x

0.5

0 90 180 270 360 x

– 0.5

– 1.0

(a) I and III (b) I and II

(c) II and III


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5.

y

y = 3 + 2sin x

0 90 180 270 360 x

–1

5

4

3

2

1

0

(ii) Max value = 5

Min value = 1

(iii) x = 270º

6.

y

2 y = cos 2x +1

1

0 90 180 270 360 x

(ii) 90, 270

(iii) 45, 135, 225 and 315

7.

y

5

y = 2 + 3 sin 2x

2

1

0 90 180 270 360 x

–1

(ii) Max value = 5

Min value = 1

(iii) 45 and 225

(iv) Quadrant II and IV

Exercise 3.4

1. 2 3 1

4

2. (i) 2 sinA cosB (ii) –2 sinA sinB 7.

56

65

8. (a) 24

25 (b)

7

25 (c)

4

3

Exercise 3.5A

1. {228.6º, 311.4º} 2. 40 ,85 ,130 ,175 ,220 ,265 ,310 ,355

3. 90 150 270 330, , , 4. 5π 11π 17π 23π

12 12 12 12

, , ,


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Exercise 3.5B

1. 2π 4π

0, , 2π3 3

,

2. 30 150 , 270, 3. π 3π 5π 7π

0, , π, , 2π4 4 4 4

, ,

4. π 3π 5π 7π

0, , π, , 2π4 4 4 4

, ,

*5. π 5π 7π 11π

, , , 6 6 6 6

Exercise 3.5C

1. π π 3π 5π 3π 7π

, , 4 2 4 4 2 4

, , ,

2. 0, 120 180 , 360,

3. 0, 22.5 67.5 , 112.5 157.5 , 180 202 5 247 5 292 5 337 5 360, , , . , . , . , . ,

4. 30 60 90 150 210 270 300 330, , , , , , ,

Exercise 3.5D

1. 60 5 193 5. , . 2. 23 5 123 9. , . 3. π

π2

,

Exercise 3.6

1. (i) π

2 (ii) 0 (iii)

π

3 (iv)

π

3 (v) does not exist (vi) –0.5 (vii)

π

6

2. (i) 3

2 (ii) 45º (iii)

1

2 3. (i) 0 (ii)

24

25

chapter 3_trigonometry and circular functions

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