1

Trigonometry Review

(I) Introduction

By convention, angles are measured from the initial line or the x-axis with respect to the origin.

If OP is rotated counter-clockwisefrom the x-axis, the angle so formed is positive.

But if OP is rotated clockwisefrom the x-axis, the angle so formed is negative.

O

P

xnegative angle

P

O xpositive angle

2

(II) Degrees & Radians

Angles are measured in degrees or radians.

rr

r1c

Given a circle with radius r, the angle subtended by an arc of length r measures 1 radian.

Care with calculator! Make sure your calculator is set to radians when you are making radian calculations.

180rad

3

(III) Definition of trigonometric ratios

r

y

hyp

oppsin

r

x

hyp

adjcos

x

y

adj

opptan

cos

sin

sin

1 cosec

cos

1sec

sin

cos

tan

1cot

x

y P(x, y)

r y

x

Note:

1sin

sin

1

Do not write cos1tan1

4

1

0 90

1

180 270 360

xy sin

Graph of y=sin x

5

1

0

1

90 180 270 360

xy cos

Graph of y=cos x

6

0 90 180 270 360

xy tan

Graph of y=tan x

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From the above definitions, the signs of sin , cos & tan in different quadrants can be obtained. These are represented in the following diagram:

All +ve sin +ve

tan +ve

1st2nd

3rd 4th

cos +ve

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What are special angles?

(IV) Trigonometrical ratios of special angles

,30 60,45

,90,0 ,180 360,270

Trigonometrical ratios of these angles are worth exploring

9

1

00sin

0sin

12

sin

02sin

12

3sin

sin 0° 0

sin 360° 0sin 180° 0

sin 90° 1 sin 270° 1

xy sin

0 2

23

1

10

10cos 1cos

02

cos

12cos

02

3cos

cos 0° 1

cos 360° 1

cos 180° 1

cos 90° 0cos 270°

1xy cos

0 2

23

1

11

00tan 0tan

undefined. is 2

tan

02tan

undefined. is 2

3tan

tan 180° 0

tan 0° 0

tan 90° is undefined tan 270° is undefined

tan 360° 0

xy tan

0 2

23

12

Using the equilateral triangle (of side length 2 units) shown on the right, the following exact values can be found.

2

3

3sin60sin

2

3

6cos30cos

2

1

6sin30sin

2

1

3cos60cos

33

tan60tan

3

1

6tan30tan

13

2

2

2

1

4sin45sin

2

2

2

1

4cos45cos

14

tan45tan

Complete the table. What do you observe?

14

15

2nd quadrant sin)sin(

cos)cos(

tan)tan(

Important properties:Important properties:

3rd quadrant sin)sin(

cos)cos(

tan)tan(

1st quadrant sin)2sin(

cos)2cos(

tan)2tan(

or 2

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Important properties:Important properties:

4th quadrant

sin)2sin( cos)2cos(

tan)2tan(or

or 2

sin)sin( cos)cos(

tan)tan(In the diagram, is acute. However, these relationships are true for all sizes of

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Complementary angles

E.g.: 30° & 60° are complementary angles.

Two angles that sum up to 90° or radians are called complementary angles.

2

2

and are complementary angles.

Recall:

2

160cos30sin

2

3

6cos

3sin

3

160cot30tan 330cot60tan

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We say that sine & cosine are complementary functions.

Also, tangent & cotangent are complementary functions.

E.g.: 50cos40sin

8

3cos

8

3tan

8cot

35cot 55tan

8sin

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E.g. 1: Simplify

(i) sin 210 (ii) cos (iii) tan(– ) (iv) sin( )

sin(180°+30)

(a) sin 210

Solution:

2

1

210° = 180°+30°

3rd quadrant

3

53

2

2

3

- sin 30 =

20

3

5cos(b) )

32cos(

2

1

2 335 4th quadrant

(c)

3

2tan

3

)3

tan(

3

cos

)3

2tan(

)

3tan(

21

sin (3 - x)

sin (2 - x)

sin ( - x)

sin x

0.6

cos (4 + x)

cos (2 + x)

0.8

cos x

Soln :

E.g. 2: If sin x = 0.6, cos x = 0.8, find

(a) sin (3 x) (b) cos (4 x).

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(V) Basic Angle

The basic angle is defined to be the positive, acute angle between the line OP & its projection on the x-axis. For any general angle, there is a basic angle associated with it.

.0or 900 So2

P

O

P

O

180° or

Let denotes the basic angle.

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)0or 900(2

360° or 2

PO

P

O

– 180°

or –

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E.g.:

P

O

55 (1st quadrant)55 basic

4

(1st quadrant)

4 basic

25

E.g.:

130 (2nd quadrant) 50130180 basic

3

2 (2nd quadrant)

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2 basic

P

O

180° or

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E.g.:

200 (3rd quadrant) 20180200 basic

4

5 (3rd quadrant)

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5 basic

P

O

– 180°

or –

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E.g.:

300 (4th quadrant) 60300360 basic

6

11 (4th quadrant)

66

112 basic

360° or 2

PO

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Principal Angle & Principal Range

Example: sinθ = 0.5

2

2

Principal range

Restricting y= sinθ inside the principal range makes it a one-one function, i.e. so that a unique θ= sin-1y exists

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E.g. 3(a): sin . Solve for θ if 2

1)

2

3( 0

4

Basic angle, α =

Since sin is positive, it is in the 1st or 2nd quadrant )2

3(

42

3

42

3 orTherefore

4

3)(

4

5 orleinadmissib

Hence, 4

3

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E.g. 3(b): cos . Solve for θ if

Since cos is negative, it is in the 2nd or 3rd quadrant )252( 0

Basic angle, α = 36.870o

870.36180252870.36180252 orTherefore

9.951.59 or

8.0)252( 0 1800

Hence, 9.951.59 or

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ry

xA

O

P(x, y)By Pythagoras’ Theorem,

222 ryx

122

r

y

r

x

(VI) 3 Important Identities

sin2 A cos2 A 1

r

xA cos

r

yA sinSince and ,

1cossin 22 AA Note:

sin 2 A (sin A)2 cos 2 A (cos A)2

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A2cos

1

(1) sin2 A + cos2 A 1

(2) tan2 A +1 sec2 A

(3) 1 + cot2 A csc2 A

tan 2 x = (tan x)2

(VI) 3 Important Identities

Dividing (1) throughout by cos2 A,

Dividing (1) throughout by sin2 A,

2)(sec A

2

cos

1

A

A2sec

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(VII) Important Formulae

(1) Compound Angle Formulae

BABABA sincoscossin)sin( BABABA sincoscossin)sin(

BABABA sinsincoscos)cos( BABABA sinsincoscos)cos(

BA

BABA

tantan1

tantan)tan(

BA

BABA

tantan1

tantan)tan(

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E.g. 4: It is given that tan A = 3. Find, without using calculator,(i) the exact value of tan , given that tan ( + A) = 5;(ii) the exact value of tan , given that sin ( + A) = 2 cos ( – A)

Solution:

(i) Given tan ( + A) 5 and tan A 3,

tan31

3tan5

3tantan155

8

1tan

A

AA

tantan1

tantan)tan(

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Solution:

sin + cos tan A = 2(cos + sin tan A)

sin + 3cos = 2(cos + 3sin )

(ii) Given sin ( + A) = 2 cos ( – A) & tan A 3,

5sin = cos

tan = 51

sin cos A + cos sin A = 2[ cos cos A + sin sin A ]

(Divide by cos A on both sides)

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(2) Double Angle Formulae

(i) sin 2A = 2 sin A cos A

(ii) cos 2A = cos2 A – sin2 A

= 2 cos2 A – 1

= 1 – 2 sin2 A

(iii)

A

AA

2tan1

tan22tan

Proof:

)sin(

2sin

AA

A

AAAA sincoscossin

AAcossin2

)cos(2cos AAA

AA 22 sincos

)cos1(cos 22 AA

1cos2 2 A

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(3) Triple Angle Formulae:

(i) cos 3A = 4 cos3 A – 3 cos A

Proof:

cos 3A = cos (2A + A)

= cos 2A cos A – sin 2A sin A

= ( 2cos2A 1)cos A – (2sin A cos A)sin A

= 2cos3A cos A – 2cos A sin2A

= 2cos3A cos A – 2cos A(1 cos2A)

= 4cos3A 3cos A

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(ii) sin 3A = 3 sin A – 4 sin3 A

Proof:

sin 3A = sin (2A + A)

= sin 2A cos A + cos 2A sin A

= (2sin A cos A )cos A + (1 – 2sin2A)sin A

= 2sin A(1 – sin2A) + sin A – 2sin3A

= 3sin A – 4sin3A

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E.g. 5: Given sin2 A & A is obtuse, find,

without using calculators, the values of

2516

(i) cos 4A (ii) sin ½A

Solution:

Since sin2 A 25

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But A is obtuse, sin A =5

45

4Asin

5

3Acos

A 5

3

4

40

(i) AA 2sin214cos 2

2

25

2421

625

527

A 5

3

4

5

3Acos

21 2(2sin cos )A A

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(ii) cos A = 1 – 2sin2 ( )2

A

5

3 = 1 – 2sin2 ( )

2

A

5

4

2sin2

A

2

Asin ( ) =

5

2

,18090 Since A .902

45 A

i.e. lies in the 1st quadrant. So 02

sin A

2

A

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E.g. 6: Prove the following identities:

(i) 1cos8cos84cos 24 AAA

Solution:

1)1cos2(2 22 A

1)1cos4cos4(2 24 AA

1cos8cos8 24 AA

(i) A4cosLHS =

= RHS

12cos2 2 A

cos 2A = cos2 A – sin2 A

= 2 cos2 A – 1

= 1 – 2 sin2 A

Recall:

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(ii) A

A

2sin

2cos1LHS =

AA

A

cossin2

)sin21(1 2

AA

A

cossin2

sin2 2

A

A

cos

sin

Atan = RHS

E.g. 6: Prove the following identities: (ii) AA

Atan

2sin

2cos1

Solution:

44

)cos1)(cos1(

)cos1)(cos1(

cos1

cos1

LHS

2

2

cos1

)cos1(

2

2

sin

)cos1(

20 where,cotcosec

cos1

cos1

E.g. 6: Prove the following identities:

(iii)

Solution:

45

sin

cos1

sin

cos1 ,

20Given (

.)1cos0 and 1sin0

2

2

sin

)cos1(

coteccos

RHS

sin

cos

sin

1

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LHS =

sin

3sinsin

cos

3coscos 33

RHS

3sin

3sinsin

cos

3coscos 33

E.g. 6: Prove the following identities:

(iv)

Solution:

sin

3sinsin

cos

3coscos 22

cossin

3cossincos3sin1

2sin

)3sin(1

21

321

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(5) The Factor Formulae (Sum or difference of similar trigo. functions)

Recall compound angles formulae:

BABABA sincoscossin)sin( ….

BABABA sincoscossin)sin( ….

BABABA cossin2)sin()sin( + :

BABABA sinsincoscos)cos( ….

BABABA sinsincoscos)cos( ….

BABABA sincos2)sin()sin( :BABABA coscos2)cos()cos( + :BABABA sinsin2)cos()cos( :

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By letting X = A + B and Y = A – B, we obtain the factor formulae:

2

cos2

sin2sinsin)1(YXYX

YX

2

sin2

cos2sinsin)2(YXYX

YX

2

cos2

cos2coscos)3(YXYX

YX

2

sin2

sin2coscos)4(YXYX

YX

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Solution:(i) LHS

= cos + cos 3 + cos 5

= cos 3 (4 cos2 – 1) = RHS

= cos 3 [ 2(2 cos2 – 1) + 1 ]

= (cos 5 + cos ) + cos 3

= 2cos 3 cos 2 + cos 3

= cos 3 [2cos2 + 1]

2

cos2

cos2

coscos Using

YXYX

YX

E.g. 8: Show that)1cos4(3cos5cos3coscos 2 (i)

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(ii)

BA

BA

coscos

sinsin

LHS =

2

cotBA

2sin

2sin2

2cos

2sin2

BABA

BABA

2sin

2cos

BA

BA

= RHS

2cot

coscos

sinsin BA

BA

BAE.g. 8: Show that (ii)

Soln:

51

(iii) LHS = sin + sin 3 + sin 5 + sin 7 = (sin 3 + sin ) + (sin 7 + sin 5 )

2

2cos

2

12sin2

2

2cos

2

4sin2

= 2sin 2 cos + 2sin 6 cos

= 2cos [ sin 6 + sin 2 ]

2

4cos

2

8sin2cos2

(iii) sin + sin 3 + sin 5 + sin 7 = 16 sin cos2 cos2 2E.g. 8: Show that

Soln:

52

= 16 sin cos2 cos2 2

= RHS

= 4 cos cos 2 sin 4

= 4 cos cos 2 [ 2 sin 2 cos 2 ]

= 8 cos cos2 2 sin 2

= 8 cos cos2 2 ( 2 sin cos )

2

4cos

2

8sin2cos2