TRIGONOMETRIC FUNCTIONGROUP MEMBERS :

NOOR AZURAH ABDULRAZAKWAN NORAZWANI MAHUSIN

IRA NUSRAT JAAFARNUR WAHIDAH SAMI’ONSITI NURHAFIZA HAFINAS

OBJECTIVES• To find the angle and convert the angle from

degree to radian or vice versa.• To recognize the trigonometric identities, sine

and cosine rule.• To solve trigonometric equations.

HISTORY• One of the oldest branches of mathematics.• Historical evidence shows that by about 1100 B.C.,

Chinese were making measurements of distance and height using right- triangle trigonometry.

• Greek astronomer Hipparcus, The Father of Trigonometry, is credited with compiling the 1st trigonometric tables.

• The trigonometry of Hipparcus and other astronomers was strictly a tool of measurement.

USES IN OUR DAILY LIFE

• Making measurements of distance and height.

• Astronomers field.

• Describing physical phenomena that are “periodic”.

ANGLES and THEIR MEASURE

• An angle is determined by rotating a ray about about its endpoint.• The starting position: initial side• The position after rotation: terminal side• The point connecting the two sides: vertex y terminal side angle initial side x vertex

Positive angles are generated Negative angles are generated

with anticlockwise rotation. with clockwise rotation.

y y

135°

x -45° x

QUADRANT‘A’ represent an angle measure. y Quadrant II Quadrant I angle: 90° <A< 180° angle: 0 <A< 90°

x Quadrant III Quadrant IV angle: 180° <A< 270° angle: 270° <A< 360°

Angles

θ θ Acute angle (0°< θ < 90°) Obtuse angle (90° <θ< 180°)

180° 90 °

Right angle ( ¼ rotation) Straight angle (1/2 rotation)

RADIAN and DEGREE

• An angle may be measured in terms of “Radians” rather than degrees.

• π radians = 180°• 2 π radians = 360°• Note: π is used to present 3.142

CONVERT: DEGREE TO RADIANS and RADIANS TO

DEGREE.

• Degree Radians• By using formula: Degree x π radians = radians 1 180°• Radians Degree• By using formula : Radians x 180° = degree 1 π radians

QUESTIONSConvert to radians.i. 60 °ii. 173°iii. 35°Convert to degree.iv. π 4ii. 7 π 8 iii. 3 π 5

SOLUTIONS• Degree to Radiansi. 60° x π radians = 1.047 radians 1 180°ii. 173° x π radians = 3.019 radians 1 180°iii. 35° x π radians = 0.611 radians 1 180°• Radians to Degreei. π x 180° = 45° 4 π radiansii. 7 π x 180° = 157.5° 8 π radiansiii. 3 π x 180° = 108 ° 5 π radians

Graph of y=sin xsin 0° sin 90° sin 180° sin 270° sin 360°

0 +1 0 -1 0

Graph y=cos xcos 0° cos 90° cos 180° cos 270° cos 360°

+1 0 -1 0 +1

Graph y=tan x•The period is π.•Graphs consists repetitions at intervals of π.•The tangent function is undefined at π/2.

RIGHT ANGLE TRIANGLE TRIGONOMETRY

Sine θ = Opposite side = y Hypotenuse Hypotenuse r r Opposite θ side ,y Cosine θ = Adjacent side = x Adjacent Hypotenuse r side , x Tangent θ = Opposite side = y Adjacent side x

• Tan θ = sin θ cos θ• Sec θ = 1 = r cos θ x• Cosec θ = 1 = r sin θ y• Cot θ = 1 = x tan θ y

TRIGONOMETRY RATIOS FOR SPECIAL

ANGLES

30° 2 45° 2 1

45°

1 60° 1

3

0° 30° 45° 60° 90°

sin θ 0

1

cos θ 1

0

tan θ 0

1 UNDEFINED

3

1

21

23

2

1

3

1

2

1

2

3

2

TRIGONOMETRIC IDENTITIES

• sin²θ + cos²θ = 1

• 1 + cot²θ = cosec²θ

• Tan²θ + 1 = sec²θ

How to proven?? y

p(x,y) phytagoras theorem:

r² = x² + y²…….① r y

θ x

From graph…

cos θ = x sin θ = y r r Divided ① by r² gives : r² = x² + y² r² r² r² 1 = x ² + y² r r

1 = cos² θ + sin ² θ……..②Divide ② by cos ² gives : 1 = cos² θ + sin²θ cos² θ cos² θ cos² θ 1 ²= 1 + sin θ ² cos θ cos θ

sec²θ = 1 + tan²θ………③Divide ② by sin²θ 1 = cos²θ + sin²θ sin²θ sin²θ sin²θ 1 ² = cos θ ² + 1 sin θ sin θ cosec ²θ = cot²θ + 1

Negative angles

• sin (-θ )= - sin θ

• cos (- θ ) = cos θ

• tan (-θ ) = - tan θ

Prove the following identities

a) ( 1 + sin θ)² = 1 + sin θ cos²θ 1- sin θ

Solution

( 1 + sin θ)² = ( 1 + sin θ ) ( 1 + sin θ ) cos²θ 1- sin²θ = ( 1 + sin θ ) ( 1 + sin θ ) ( 1 - sin θ ) ( 1 + sin θ )

= 1 + sin θ 1- sin θ

b) ( 1 + tan² θ )² = sec ⁵ θ cos θ

solution

( 1 + tan² θ )² = ( sec²θ )² cos θ cos θ = sec⁴ θ cos θ = 1 x sec⁴ θ cos θ = sec θ x sec⁴ θ = sec ⁵ θ

c) ( sin θ + cos θ )² + ( sin θ - cos θ )² = 2SolutionLHS= ( sin θ + cos θ )² + ( sin θ – cos θ) ( sin θ – cos θ)= sin²θ + 2 sin θ cos θ + cos²θ + sin²θ – 2 sin θ cos θ +

cos²θ= sin² θ + cos² θ + sin² θ + cos² θ = 1+ 1= 2LHS = RHS SO, PROVEN.

d) sec θ – tan θ = cos θ 1 + sin θSolutionRHS, cos θ = cos θ 1 – sin θ 1 + sin θ 1 – sin θ = cos θ –cos θ sin θ 1- sin ²θ

= cos θ –cos θ sin θ cos²θ = cos θ - cos θ sin θ cos²θ cos²θ = 1 - sin θ cos θ cos θ = sec θ - tan θ RHS = LHS , SO PROVEN

Trigonometric Equation

• A trigonometric equation is an equations that contains a trigonometric expression with a variable, such as sin x

Step in solving trigonometric equations

• Step 1 : Identify the range for the given angle• Step 2 : identify the quadrant for the basic

angle• Step 3 : Find the basic angle (α )• List all the answers in radian or degree ( depends on the given range )

Solve the following equations for angles in the given range

a) tan θ = 1 , 0 ≤ � θ ≤ 360

b) tan 2x = 1 0 ≤ x ≤ 360 �

�

solutions

a) Step 1 : 0 ≤ � θ ≤ 360 Step 2 : quadrant 1 and 3Step 3 : tan α = 1

α = = 45 �

Step 4 : θ = 45 , 225 � �

1tan 1

b) tan 2x = 1step 1 : 0 ≤ � θ ≤ 360

0 ≤ 2x ≤ 720 �step 2 : quadrant 1 and 3step 3 : tan α = 1

α = α = 45 �

step 4 : 2x = 45 , 225 , 405 ,585 � � � � x = 22.5 , 112.5 , 202.5 , 292.5 � � � �

1tan 1

TRIGONOMETRIC EQUATION

1. Solution of trigonometric equation such as = k , = k , = k

2. Solve equations in quadratic form3. Express , and in

term of t where sin cos tan

tan2

t

sin cos tan

Express , & in term of t where

2t 1+t²

1-t²

sin cos tan tan2

t

2

2

2 tantan 2

1 tan

2 tan2tan

1 tan2

22 2 2

2 2 2 4

2 4 2

2 2 2

22 2

2

(2 ) (1 )

4 1 2

2 1

1 1

1

1

x t t

x t t t

x t t

x t t

x t

x t

2

2

1

t

t

2t 1+t²

2

2

2

2

2tan

12

sin1

1cos

1

t

tt

t

t

t

Equation in the form a cos Ɵ + b sin Ɵ =k

Can be solved using these expression

21 t

Example• Solve the equation 3cos x -8sin x= -2, 0°≤Ɵ≤360°

2

2 2

2 2

2 2

2

2

2

3cos 8sin 2

1 23 8 2

1 1

3 1 8 2 2 1

3 3 16 2 2

16 5 0

4

2

16 16 4 1 5

2

16 236

20.3066

16.3066

x x

t t

t t

t t t

t t t

t t

b b act

a

t

t

tan2

xt

0 360

0 1802

x

x

1

tan 0.30662

tan 0.30662

17.052

34.1

x

x

x

x

1

tan 16.30662

tan 16.30662

86.49

180 86.492

93.512

187.02

x

x

x

x

x

34.9 ,187.02x

Tan positive in quadrant 1 and 3 tan negative in quadrant 2 and 4

Express a cos Ɵ ± b sin Ɵ as R cos (Ɵ±α) or R sin (Ɵ±α)

Equating the coefficient of cos Ɵ: R cos α = a …………….(1)Equating the coefficient of sin Ɵ: R sin α = b …………….(2)

(1)²+(2)²

cos cos sin

(cos cos sin sin ) cos sin

cos cos sin sin cos sin

R a b

R a b

R R a b

2 2 2 2 2 2

2 2 2 2 2

2 2 2

2 2

cos sin

cos sin

R R a b

R a b

R a b

R a b

(1)÷(2)

where where

sin

cos

tan

R b

R ab

a

cos sin cosa b R

2 2

tan

R a b

b

a

sin cos sin( )a b R

2 2

tan

R a b

b

a

Example Express 4 cos Ɵ – 3 sin Ɵ = 1 in the form of R cos (Ɵ + α) andsolve for Ɵ.

4cos 3sin cos( )

(cos cos sin sin )

cos cos sin sin

cos 4

sin 3

R

R

R R

R

R

2 24 3

25

5

R

R

R

1

3tan

43

tan4

36.87

cos( 36.87 )R

1

4cos 3sin 5cos 36.87

4cos 3sin 1

5cos 36.87 1

1cos 36.87

51

( 36.87 ) cos5

36.87 78.46

36.87 78.46 ,360 78.46

41.59 ,244.67

Equation in linear form

Example 1Solve 4 sin θ – 3 cos θ = 0 for angles in therange Solution

4 sin θ = 3 cos θ

=

sin

cos

3

4

=

tan α = α = α = 36.9 � θ = 36.9 , 216.9 � �

tan 3

4

3

4

1 3tan

4

Equation in quadratic form

Solve the following trigonometric equations1. 2 sin ² x+ 5 cos x + 1 for -180 ≤ x ≤ 180 � �Solution

sin ² x + cos ² x = 1sin ² x = 1- cos ² x2(1- cos ² x) + 5 cos x + 1 = 02 - 2 cos ² x + 5 cos x + 1 = 0 - 2 cos ² x + 5 cos x + 3 = 0

2 cos ² x – 5 cos ² x – 3 = 0let y = cos x2y²- 5y – 3 = 0( y-3 )( 2y+1 ) = 0y = 3 and y = cos x = cos α =

α = α = 60 �

1

2

1

21

21 1

cos2

x = 120 , -120 � �x = -120 , 120 � �

cos x = 3 ( no solution )

2) 3 cot ²θ + 5 cosec θ + 1 for -2 ≤θ≤ 2 solution

3 ( cosec ²θ-1) + 5 cosec θ + 1 = 03 cosec ²θ – 3 + 5 cosec θ + 1 = 03 cosec ²θ + 5 cosec θ – 2 = 0

let y = cosec θ3y² + 5y – 2 = 0( 3y – 1 )( y + 2 ) = 0y = and y = - 2 cosec θ = =

= 3 ( no solution ) cosec θ = -2

1

3 1

31

sin1

3

sin

= -2 = - 2

sin α = sin α =

α = 30 �α =

θ = , ,

1

sinsin

1

2

1

2

6

7

6

6

5

6

sin sin sin

cos cos cos

tan tan tan

x y x y

x y x y

x y x y

COMPOUND ANGLEusing substitution, it is clear to see that;

example3

sin(30 30 ) sin 602

1 1 3sin 30 sin 30 1

2 2 2

SUM & DIFFERENCE OF SINE

• Replacing y with –y and nothing that• Cos(-y)=cos y since cosine is even function• Sin(-y)=-sin y since sine is odd functionsin( )x y sin cos cos sinx y x y

sin sin sinx y x y

sin cos cos sinx y x y

sin cos cos sinx y x y

Example

• Find the exact value of sin105 sin105 sin 60 45

sin 60 cos 45 cos 60 sin 45

3 2 1 2

2 2 2 2

6 2

4

SUM & DIFFERENCE OF COSINE

cos( ) cos cos sin sinx y x y x y

cos cosx y x y

cos cos sin sinx y x y

cos cos sin sinx y x y

Example

• Find the exact value of cos15º cos15 cos 60 45

cos60 cos 45 sin 60 sin 45

1 2 3 2

2 2 2 2

2 6

4

SUM & DIFFERENCE OF TANGENT

• As we know…sin

tancos

sintan

cos

x yx y

x y

sin cos cos sin

cos cos sin sin

x y x y

x y x y

sin cos cos sincos cos cos cos

tan( )cos cos sin sincos cos cos cos

x y x yx y x y

x yx y x yx y x y

sin sincos cos

sin sin1

cos cos

x yx y

x yx y

tan tan

1 tan tan

x y

x y

Example• Find the value of 75º in exact radical form.Solution…

75º=45º+30º let x=45º y=30º

tan tantan( )

1 tan tan

x yx y

x y

tan 45 tan 30tan(45 30 )

1 tan 45 tan 30

1

131

1 13

3 1

3 1

2 3

COFUNCTION FORMULAS

•In a right triangle, the two acute angles are complementary. •Thus, if one acute angle of a right triangle is x, the other is

90 x

cos cos cos sin sinx y x y x y

cos cos cos sin sin2 2 2

y y y

0 cos 1 sin

sin

y y

y

cos sin2

y y

The cofunction identity for cosine

2y x

let

• The cofunction identity for cosine

cos sin2 2 2

x x

cos sin2

x x

sin cos2

x x

sin cos2

x x

• Divide all equation with cos2

x

sincos2

cos cos2 2

xx

x x

sincos2sincos

2

xx

xx

tan cot2

x x

The cofunction identity for tangent

B

a c

C A b

asin A=

c

acos B =

c

atan A = b

acot A = b

csec A =

ac

csc B = a

90A B

EXAMPLE…

Write in term of its cofunction• Sin11º = cos (90º-11º) = cos79º

• Cot 87º = tan (90º-87º) = tan 3º

• Sec 52º =csc (90º-52º) =csc 38º

DOUBLE ANGLE FORMULAE…

• sin 2x = 2 sin(x) cos(x)

• cos 2x = = =

• tan 2x =

2cos sin 2x x22cos 1x

21 2sin x

2

2 tan

1 tan

x

x

DOUBLE ANGLE…

We know that,

If we let B=A,then

Hence,

sin(A+B)=sinAcosB+sinBcosA

sin(A+A)=sinAcosA+cosAsinA

sin2A=2sinAcosA

sin 2 2sin cosA A A

We know that,

If we let B=A,then

Hence,

2 2cos2 cos sinA A A

cos( ) cos cos sin sinA B A B A B

cos( ) cos cos sin sinA A A A A A 2 2cos2 cos sinA A A

2cos2 2cos 1A A 2cos 2 1 2sinA A

We know that

If we let B=A,

Hence,

2

2 tantan 2

1 tan

AA

A

tan tantan( )

1 tan tan

A BA B

A B

tan tantan( )

1 tan tan

A AA A

A A

2

2 tantan 2

1 tan

AA

A

Example 1…If and lies in quadrant II, find the exactvalue of .Solution:

5sin

3

sin 2

5sin

3

y

r

2 2 2

2

2

5 13

25 169

144

x

x

x

144 12x

So,

12cos

13

x

r

sin 2 2sin cos

5 12 120sin 2 2

13 13 169

EXAMPLE 2….

with is acute angle, find the exact value of: Solution:a)

3if tan

4

a) tan 2

2

2 tantan 2

1 tan

32

43 3

14 4

24

7

b) tan 4

Solution: b) tan 4

tan 4 tan(2 2 ) tan 2 tan 2

1 tan 2 tan 2

2

2 tan 2

1 tan 2

2

242

7

241

7

336

527

HALF-ANGLE FORMULAE….

sin 2sin cos2 2

2 2cos cos sin2 2

22cos 12

21 2sin2

2

2 tan2tan

1 tan2

HALF-ANGLE….

We know that,

Let ,

Hence,

sin 2 2sin cosA A A

2A

sin 2sin cos2 2

sin 2 2sin cos2 2 2

sin 2sin cos2 2

We know that,

Let , Hence,

2 2cos cos sin2 2

2 2cos 2 cos sinA A A 2cos 2 2cos 1A A

2cos 2 1 2sinA A

2A

2 2cos 2 cos sin2 2 2

2

2

2cos 12

1 2sin2

2 2cos cos sin2 2

22cos 12

21 2sin2

We know that,

Letting ,

Hence,

2

2 tan2tan

1 tan2

2

2 tantan 2

1 tan

AA

A

2A

2

2 tan 22tan 2

2 1 tan2

2

2 tan2tan

1 tan2

Example….Without using calculator,compute the exact valueof cos 112.5⁰.Solution:cos 112.5⁰= cos

112.5⁰ lies in quadrant II,where only the sine andcosecant are (+)Thus, - sign is used in the half-angle formulae

225

2

o

cos 112.5⁰= cos

225

2

o

01 cos225

2

21

2

2

2 2

4

2 2

2

THE LAW OF SINES

If A, B, and C are the measures of the angles of a triangle, and a, b, and c are the lengths of the sides opposite these angles, then

A

a

sin=

B

b

sin=

C

c

sin

The ratio of the length of the side of any triangle to the sine of the angle opposite that side is the same for all three sides of the triangle.

EXAMPLE

Solve triangle ABC if A = 50°, C = 33.5°, and b = 76.

= 50°,

C

BA50°

33.5°b = 76

a

c

THE LAW OF COSINES

If A, B and C are the measures of the angles of a triangle, and a, b and c are the lengths of the sides opposite these angles, then

a2 = b2 + c2 – 2bc cos A b2 = a2 + c2 – 2ac cos B c2 = a2 + b2 – 2ab cos C

The square of a side of a triangle equals the sum of the squares of the other two sides minus twice their product times the cosine of their included angle

EXAMPLE

• = 50°,

Solve the triangle with A = 60°, b = 20, and c = 30.

C

A Bc = 30

b = 20

AREA OF TRIANGLE

Area = 1/2(a)(b)(SinC)

EXAMPLE

Find the area of this triangle

6cm

52°

14cm