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Term 3 : Unit 1 Trigonometric Functions Name : ____________ ( ) Class : _____ Date : _____ 1.1 Trigonometric Ratios and General Angles 1.2 Trigonometric Ratios of Any Angles

# Term 3 : Unit 1 Trigonometric Functions Name : ____________ ( ) Class : _____ Date : _____ 1.1 Trigonometric Ratios and General Angles 1.2 Trigonometric

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Term 3 : Unit 1Trigonometric Functions

Name : ____________ ( ) Class : _____ Date : _____

1.1 Trigonometric Ratios and General Angles

1.2 Trigonometric Ratios of Any Angles

Trigonometric Equations

Objectives

1.1 Trigonometric Ratios and General Angles

In this lesson, we will learn how to find the trigonometric ratios for

acute angles, particularly those for 30°, 45° and 60° (or

,3

and 4

,6

Trigonometric Ratios of Acute Angles

The three trigonometric ratios

are defined as

OPQ is a right angled triangle

Trigonometric Equations

oppositehypotenuse

sin

cos

tan

oppositeoppositehypotenusehypotenuse

PQ

OP

OQ

OP

PQ

OQ

Example 1

In the right-angled triangle ABC, tan θ = 2. Find sin θ and cos θ.

Solution

Trigonometric Equations

A B

C

θ

Since tan θ = ,1

2 2

1

BC = 2 units and AB = 1 unit.

By Pythagoras’ Theorem, AC = . units 5

5

AC

BCsin

5

2

AC

ABcos

5

1

Trigonometric Ratios of Special Angles

Draw a diagonal to the

square.

Draw a unit square.

Trigonometric Equations

The length of the diagonal is √2 and the angle is 45°.

1sin 45

2

1cos 45

2

tan 45 1

Trigonometric Ratios of Special Angles

Draw an equilateral triangle of side 2 cm.

Trigonometric Equations

The altitude bisects the base of the

triangle.

3sin 60

2

1cos 60

2

tan 60 3

Draw an altitude.

The length of the altitude is √3 and the angles are 60° and

30°.

3cos30

2

1sin 30

2

1tan 30

3

Trigonometric Ratios of Complementary Angles

In the right-angled triangle OPQ

but OPQ = 90°– θ

Trigonometric Equations

cosOQ

OP

sinPQ

OP

tanPQ

OQ

sin 90

cos 90

1

tan 90

OPQ = – θ

sin2

cos2

1

tan2

2

Trigonometric Equations

Example 2

Using the right-angled triangle in the diagram, show that sin(900 – θ) = cos θ. Hence, deduce the value of

SolutionP Q

R

θ

a

bc

900 – θ.70sin20cos

20cos

.)sin(90 and cos,In c

a

c

aPQR

Thus, sin(900 – θ) = cos θ.

sin 700 = sin (900 – 200)

= cos 200

2

1

20cos2

20cos

70sin20cos

20cos

Trigonometric Equations

O

y

x

P

O

y

x

P

180° –

P '

O

y

x

P

180° +

P '

O

y

x

P

360° –

P '

O

y

x

P

O

y

x

P

P '

O

y

x

P

– 180

P '

O

y

x

P

– – 180

P '

Consider angles in the Cartesian plane.

OP is rotated in an anticlockwise direction around the origin O. The basic (reference) angle that OP makes with the positive x–axis is α.

Now OP is rotated in the clockwise direction.

O

y

x

P

– 360

220

180

Example 3

Given that 00 < θ < 3600 and the basic angle for θ is 400, find the value of θ if it lies in the (a) 3rd quadrant, (b) 4th quadrant.

Solution(a) (b)

320

360

Trigonometric Equations

Using the complementary angle identity.

Substitute for sin θ.

Trigonometric Equations

cos 90 sin

1Given that sin , find the value of sin cos 90

2

sin cos 90 sin sin 1 1 1

2 2 4

1tan 90

tanA

A

Given that tan 2, find the value of 2 tan tan 90A A A

1 12 4

2 22

12 tan tan 90 2 tan

tanA A A

A

Using the complementar

y angle identity.

Substitute for tan A.

Example 4

Trigonometric Equations

1

2sin 45

cos30 sin 60 3 3

2 2

sin 45Without using a calculator, evaluate

cos30 sin 60

1 1

2 3 6

Example 5

Solution

Trigonometric Equations

70 , 180 70 , 180 70 , 360 70

Find all the angles between 0° and 360° which make a basic angle of 70°.

The angles are as follows:

70 , 110 , 250 , 290

Example 6

Solution

Trigonometric Equations

Objectives

In this lesson, we will learn how to

• extend the definitions of sine, cosine and tangent to any angle,

• determine the sign of a trigonometric ratio of an angle in a

• relate the trigonometric functions of any angle to that of its basic

(reference) angle and solve simple trigonometric equations.

1.2 Trigonometric Ratios of Any Angles

Trigonometric Ratios of Any AnglesThe three

trigonometric ratios are defined as

Trigonometric Equations

x

yr

sin

cos

tan

yyr

x

r

x

PQ

OP

OQ

OP

PQ

OQ

PQ = yOQ = x

22 yxrOP

Example 7

Find the values of cos θ, sin θ and tan θ when θ = 1350.

Solution

Trigonometric Equations

When θ = 1350, 1800 – θ = 450.(basic angle)

P has coordinates (1, -1)

and

2

1)1( 22

r

Trigonometric Equations

2

1135cos

2

1135sin

11

1135cos

x

y

O

P(a, b)

rb

Signs of Trigonometric Ratios in Quadrants

Trigonometric Equations

x

y

O

P( – a, b)

rb

– a

θ = α

sinb

r

cosa

r

tanb

a

sin

cos

tan

P has coordinates ( a, b )

x

y

O

P( – a, – b)

r– b

– a

θ = ( 180° – α )

P has coordinates ( – a, b )

cosa

r

cos

tanb

a

tan

sinb

r

sin

tanb

a

x

y

O

P(a, – b)

r– b

a

θ = ( 180° + α )

P has coordinates ( – a, – b )

θ = ( 360° – α )

P has coordinates ( a, – b ).

tanb

a

Trigonometric Equations

902

For positive ratios

Signs of Trigonometric Ratios in Quadrants

3270

2

180 0 0 360 2

In the four quadrantsS (sin θ) A ( all )

T (tan θ) C (cos θ)The signs are

summarised in this diagram.

x

y

O

120°

Trigonometric Equations

Example 8

Without using a calculator, evaluate cos 120°.

Solution

Basic angle,

120° is in the 2nd quadrant, so cosine is negative

180 120

60

cos120 cos60 1

2

AS

T C

Trigonometric Equations

Example 9

Find all the values of θ between 0° and 360° such that

sin θ = – 0.5.

Solution

For the basic angle,

Since sin θ < 0, θ is in the 3rd or 4th quadrant,

sin 0.5 30

180 or 360

180 30 or 360 30

x

y

O

AS

T C

Basic Trigonometric Equations

210 or 330

Trigonometric Equations

For the basic angle,

θ is in the 1st, 2nd, 3rd or 4th quadrant,

1sin

2 045

x

y

O

AS

T C

22sin 1 0 2 1sin

2

1sin

2

45 ,180 45 ,180 45 ,360 45

45 ,135 ,225 ,315

o o o o o o o

o o o o

3 5 7

, , or 4 4 4 4

Example 10

Find all the values of θ between 0o and 360o such that

2sin2 θ – 1 = 0.

Solution