Trigonometric function

TRIGONOMETRIC FUNCTION
GROUP MEMBERS :
NOOR AZURAH ABDULRAZAK
WAN NORAZWANI MAHUSIN
IRA NUSRAT JAAFAR
NUR WAHIDAH SAMI’ON
SITI 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

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

SOLUTIONS
• Degree to Radians
i. 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 Degree
i. π x 180° = 45°
4 π radians
ii. 7 π x 180° = 157.5°
8 π radians
iii. 3 π x 180° = 108 °
5 π radians

Graph of y=sin x
sin 0° sin 90° sin 180° sin 270° sin 360°
0 +1 0 -1 0

Graph y=cos x
cos 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°
3
1
60°

1

0° 30° 45° 60° 90°
sin θ 0 1 1
1 3
2 2 2
cos θ 1 1 1 0
3
2 2 2
tan θ 0 1 1 UNDEFINED
3
3

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……..②
Divide 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………③
Divide 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 = 2
Solution
LHS
= ( 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
=2
LHS = RHS SO, PROVEN.

d) sec θ – tan θ = cos θ
1 + sin θ
Solution
RHS, 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 3
Step 3 : tan α = 1
1
α = tan 1
= 45 ̊
Step 4 : θ = 45 ̊ , 225 ̊

b) tan 2x = 1
step 1 : 0 ̊ ≤ θ ≤ 360
0 ̊ ≤ 2x ≤ 720
step 2 : quadrant 1 and 3
step 3 : tan α = 1
α= tan 1 1
α = 45 ̊
step 4 : 2x = 45 ̊, 225 ̊, 405 ̊,585 ̊
x = 22.5 ̊, 112.5 ̊, 202.5 ̊, 292.5 ̊

TRIGONOMETRIC EQUATION
1. Solution of trigonometric equation
such as sin = k , cos = k ,tan = k
2. Solve equations in quadratic form
3. Express sin , cos and tan in
term of t where t tan
2

Express sin , cos & tan in term of t where t tan
2
2 tan
tan 2
1 tan 2
2 tan 2t
tan 2
1 tan 2 1 t2
2
2
2
x (2t ) (1 t 2 ) 2 2t 1+t²

x2 4t 2 1 2t 2 t4
x2 t4 2t 2 1
2 2 2 1-t²
x t 1 t 1
2 2 2
x t 1
x t2 1

2t
tan
1 t2
2t
sin
1 t2
1 t2
cos 2t 1+t²
1 t2

Equation in the form a cos
Ɵ + b sin Ɵ =k 1 t2
Can be solved using these
expression

Example
• Solve the equation 3cos x -8sin x= -2, 0°≤Ɵ≤360°
3cos x 8sin x 2
1 t2 2t
3 8 2
1 t2 1 t2
3 1 t2 8 2t 2 1 t2
3 3t 2 16t 2 2t 2
t 2 16t 5 0
b b 2 4ac
t
2a
16 162 4 1 5
2
16 236
2 x
t 0.3066 t tan
2
t 16.3066

0 x 360
x
0 180
2

Tan positive in quadrant 1 and 3 tan negative in quadrant 2 and 4
x
tan 16.3066
x 2
tan 0.3066 x 1
2 tan 16.3066
x 2
tan 1 0.3066 86.49
2
x x
17.05 180 86.49
2 2
x 34.1 x
93.51
2
x 187.02

x 34.9 ,187.02

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

R cos a cos b sin
R(cos cos sin sin ) a cos b sin
R cos cos R sin sin a cos b sin

Equating the coefficient of cos Ɵ: R cos α = a …………….(1)
Equating the coefficient of sin Ɵ: R sin α = b …………….(2)
R 2 cos 2 R 2 sin 2 a 2 b2
(1)²+(2)² R 2 cos 2 sin 2 a 2 b2
R2 a 2 b2
R a 2 b2

R sin b
(1)÷(2) R cos a
b
tan
a

a cos b sin R cos  a sin b cos R sin( )

where where

R a 2 b2 R a 2 b2
b b
tan tan
a a

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

R 42 32 3
tan
4
R 25
1 3
R 5 tan
4
R cos( 36.87 )
36.87

4 cos 3sin 5cos 36.87
4 cos 3sin 1
5cos 36.87 1
1
cos 36.87
5
1 1
( 36.87 ) cos
5
36.87 78.46
36.87 78.46 ,360 78.46
41.59 , 244.67

Equation in linear form
Example 1
Solve 4 sin θ – 3 cos θ = 0 for angles in the
range
Solution
4 sin θ = 3 cos θ

sin 3
= 4
cos

3
tan = 4

3
tan α =
4
1 3
α = tan
4
α = 36.9 ̊
θ = 36.9 ̊, 216.9 ̊

Equation in quadratic form
Solve the following trigonometric equations
1. 2 sin ² x+ 5 cos x + 1 for -180 ̊≤ x ≤ 180 ̊
Solution
sin ² x + cos ² x = 1
sin ² x = 1- cos ² x
2(1- cos ² x) + 5 cos x + 1 = 0
2 - 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 = 0
let y = cos x
2y²- 5y – 3 = 0
( y-3 )( 2y+1 ) = 0
1
y = 3 and y = 2
cos x = 1
2
1
cos α = 2
1 1
α = cos
2
α = 60 ̊

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 = 0
3 cosec ²θ – 3 + 5 cosec θ + 1 = 0
3 cosec ²θ + 5 cosec θ – 2 = 0

let y = cosec θ
3y² + 5y – 2 = 0
( 3y – 1 )( y + 2 ) = 0
1
y= and y = - 2
3
1
cosec θ =
1 3
=1
sin 3

sin = 3 ( no solution )
cosec θ = -2

1 = -2
sin
sin = - 2
1
sin α = 2
sin α = 12
α = 30 ̊
α=6

7 5
θ= , ,
6 6 6

COMPOUND ANGLE
using substitution, it is clear to see that;
sin x y sin x sin y
cos x y cos x cos y
tan x y tan x tan y
example
3
sin(30 30 ) sin 60
2
1 1 3
sin 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 function
sin( x y) sin x cos y cos x sin y
sin x y sin sin x y
sin x cos y cos x sin 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( x y) cos x cos y sin x sin y

cos x y cos x y
cos x cos y sin x sin y

Example
• Find the exact value of cos15º
cos15 cos 60 45
cos 60 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
tan
cos

sin x y
tan x y
cos x y
cos x cos y  sin x sin y

cos x cos y cos x cos y
tan( x y)

cos x cos y cos x cos y

sin x sin y
cos x cos y
sin x sin y
1 
cos x cos y

tan x tan y
1  tan x tan y

Example
• Find the value of 75º in exact radical form.
Solution…
75º=45º+30º let x=45º y=30º
tan x tan y
tan( x y)
1 tan x tan y
tan 45 tan 30
tan(45 30 )
1 tan 45 tan 30
1
1
3
1
1 1
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 x y cos x cos y sin x sin y

cos y cos cos y sin sin y
2 2 2
0 cos y 1 sin y
sin y
The cofunction identity for cosine

cos y sin y
2

let y
2
x

cos x sin x
2 2 2

cos x sin x
2

• The cofunction identity for cosine

sin x cos x
2

sin x cos x
2
cos x
• Divide all equation with 2
sin x
2 cos x
cos x cos x
2 2

sin x
2 cos x
sin x
cos x
2

The cofunction identity tan x cot x
for tangent 2

B A B 90
a a
sin A= cos B =
c c

a a
tan A = cot A =
a c b b
c c
sec A = csc B =
a a
C A
b

Write in term of its cofunction
• Sin11º
= cos (90º-11º) • Sec 52º
= cos79º =csc (90º-52º)
=csc 38º
• Cot 87º
= tan (90º-87º)
= tan 3º

• sin 2x = 2 sin(x) cos(x)

2
• cos 2x =cos x sin 2 x
2
= 2cos x 1
= 1 2sin 2 x

2 tan x
• tan 2x =
1 tan 2 x

sin 2 A 2sin Acos A

We know that,
sin(A+B)=sinAcosB+sinBcosA

If we let B=A,then
sin(A+A)=sinAcosA+cosAsinA
Hence, sin2A=2sinAcosA

2 2
cos2 A cos A sin A
We know that,
cos( A B) cos A cos B sin A sin B
If we let B=A,then
cos( A A) cos A cos A sin Asin A
2 2
Hence, cos2 A cos A sin A
2
cos 2 A 2cos A 1
2
cos 2 A 1 2sin A

2 tan A
tan 2 A
1 tan 2 A

We know that
tan A tan B
tan( A B)
1 tan A tan B
If we let B=A,
tan A tan A
tan( A A)
1 tan A tan A
2 tan A
Hence, tan 2 A
1 tan 2 A

Example 1…
5
If sin andlies in quadrant II, find the exact
3
value of sin 2 .
Solution:
5 y
sin
3 r
2 2 2
x 5 13
2
x 25 169
2
x 144
x 144 12

So, x 12
cos
r 13
sin2 2sin cos
5 12 120
sin 2 2
13 13 169

EXAMPLE 2….
3
if tan with is acute angle, find the
4
exact value of:
a) tan 2 b) tan 4
2 tan
Solution:a) tan 2 2
1 tan
3
2
4
3 3
1
4 4
24
7

b) tan 4
Solution:
tan 4 tan(2 2 )
tan 2 tan 2
1 tan 2 tan 2
2 tan 2
2
1 tan 2
24
2
7
2
24
1
7

336
527

HALF-ANGLE
FORMULAE….
 sin 2sin cos
2 2

 cos cos 2
2
sin 2
2
2cos 2 1
2
1 2 sin 2
2
2 tan
 tan
2
1 tan 2
2

HALF-ANGLE….
sin 2sin cos
2 2

We know that, sin 2 A 2sin Acos A

Let A ,
2

sin 2 2sin cos
2 2 2

Hence, sin 2sin cos
2 2

cos cos 2 sin 2
2 2

We know that,
cos 2 A cos 2 A sin 2 A
cos 2 A 2cos 2 A 1
cos 2 A 1 2sin 2 A
Let A , Hence,
2
cos 2 cos 2
sin 2 cos cos 2 sin 2
2 2 2 2 2
2 cos 2 1 2cos 2 1
2 2
1 2sin 2 1 2 sin 2
2 2

2 tan
2
tan
1 tan 2
2

2 tan A
We know that,tan 2 A
1 tan 2 A

Letting A ,
2
2 tan 2
2
tan 2
2
1 tan 2
2
Hence,
2 tan
2
tan
1 tan 2
2

ExAMPLE….
Without using calculator,compute the exact value
of cos 112.5⁰.
Solution:
o
cos 112.5⁰= cos 225
2
112.5⁰ lies in quadrant II,where only the sine and
cosecant are (+)
Thus, - sign is used in the half-angle formulae

cos 112.5⁰= cos 225o
2
1 cos2250
2
2
1
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 b c
= =
sin A sin B sin C
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.

C

33.5°
b = 76
a

= 50°,
50°
A B
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
Solve the triangle with A = 60°, b = 20, and c = 30.

C

b = 20

• = 50°,
A B
c = 30

AREA OF TRIANGLE

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

EXAMPLE
Find the area of this triangle

6cm

52°

14cm

Trigonometric function

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