FORM IV MATHS

CHAPTER FOUR

TROGONOMETRY III
In form iv we defined trigonometric ratios as relationship between one
angle(size) and the lengths of two sides in a right-angled triangle.
Given an angle Ө in a right-angled triangle ABC then:

b
C

C B
a

Side opposite to ¿ A = a
Side opposite to ¿ B = b
Side opposite to ¿ C = c
In relation to angle Ө
C-hypotenuse side
a – adjacent side
b – opposite side
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Hence Cos Ө = ad ja ce nt
h y pot e nu se

cos Ө = a
c

Sin Ө = o p po si t e
h y pot e nu se

= b
c

Tan Ө = o p p osi t e
ad j ace nt

=c
a

Tan Ө = sin ⁡Ө
cos⁡Ө

In form three we were able to get the trigonometric ratios of angles

greater than 900 using a unit circle
A unit circle has four quadrants

1st
2nd

A
S

T C
3rd 4th

A – All the trigonometric ratios are positive

S- Sine is positive in 2nd quadrant
T – Tangent is positive in 3rd quadrant
C – Cosine is positive in 4th quadrant
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Example
Find the trigonometric ratio of
a) Sin 150
b) Cos 240
c) Cos 300
Solution
a) Sin 150 = sin (180 – 150) 150- 2nd Quadrant
=Sin 300 Sine is positive in 2nd Quadrant
240 – 3rd Quadrant
= 0.5
Cos is –ve on 3rd Quadrant.
b) Cos 240 = -cos (180 + 60)
= - cos 60

= - 0.5
th
c) Cos 300 = cos (360 – 300)Cos is +ve on 4 Quadrant

300 is in 4th Quadrant = cos 60

= 0.5

Deriving the relation

Cos 2x + sin2x = 1
Using a unit circle (circle radius one unit) and point A on the unit circle.

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 A

Joining A to the centre O and forming a right-angle triangle OAB

¿ AOB is Ө then using the Pythagoras theorem

OB2 + AB2 = OA2
But sin Ө = AB and OA = r
OA

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Sin Ө = AB multiply both sides by r
r

AB = r sin Ө
Cos Ө = OB OA = r
OA

Cos Ө = OB multiply both sides by r

r

OB = r Cos Ө
Substituting the values of OB with r cos Ө and AB with r sin Ө in the
OB2 + AB2 = OA2
(r Cos Ө)2 + (r sin Ө)2 = r2
Opening the brackets
R2 cos2Ө + r2sin2Ө = r2
Dividing each part by r2
2 2 2 2 2
r cos Ө
2
+ r s in Ө
2
= r
2
r r r

Cos2Ө + sin 2Ө = 1
Also, tan Ө = o p p osi t e = AB
ad j ace nt OB

But AB = r sin Ө and OB = r Cos Ө

Tan Ө = r sin ⁡Ө substitution
rcos ⁡Ө

Tan Ө = sin ⁡Ө
cos⁡Ө

Which means squaring both sides

(Tan Ө)2 = ( sin ⁡Ө )2
cos⁡Ө

Tan2Ө = sin Ө
2
2

co s Ө

Proving identities
Example 1

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Prove the identity
a) 1 + sin ⁡x = 1
tan ⁡x cos⁡x cos⁡x sin ⁡x

1 + sin ⁡x is on the left-hand side (LHS)

tan ⁡x cos⁡x

And
1 is on the Right-hand side (RHS)
cos⁡x sin ⁡x

To prove, we show that after working the RSH is equal to the LSH.
1 + sin ⁡x = cos ⁡x + sin ⁡x tan x = sin ⁡x
t an x cos⁡x sin ⁡x cos⁡x cos⁡x
1
1 = 1÷ SI N X = C OS X
sin ⁡x
tan ⁡x C OS X SI N X
cos ⁡x

Taking LCM if sin x and cos x = sin x cos x

And multiply by the LCM
2 2
= C OS X+ Si n x
s i n x cos ⁡x

Since cos2x + sin 2x = 1

Substituting cos2x + sin2x with = 1
si n x cos ⁡x

= 1
cos ⁡x sin ⁡x

Hence the LHS is now identical with the RHS

1 + sin ⁡x = 1
∴
T an x cos⁡x cos⁡x sin ⁡x

b) Cos4Ө – sin4Ө = cos2Ө – sin2Ө

Cos 4Ө – sin4Ө = (cos2Ө – sin2Ө)2 using difference of two
squares
= (cos2Ө – sin2Ө) (cos2Ө + sin2Ө)
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 But cos 2Ө + sin2Ө = 1
= (cos2Ө – sin2Ө) (1) substituting with 1
= cos2Ө – sin2Ө

4Ө – sin4Ө = cos2Ө – sin2 Ө

∴ Cos

Exercise
1. Prove the identity
2 3
s i n Ө cos ⁡Ө+ c o s Ө = 1
sin ⁡Ө x

More exercise
KLB book 4 page 92 – 93
Advancing mathematic book 4 page 55

WAVES
Amplitude and period
Amplitude – the distance the curve covers below or above the x – axis.
Period – the interval at which a wave repeats itself (one complete cycle)

Sine graph (y = sin x)

Taking the value of x between 00 and 7200 at interval 900.
A table is formed as follows;
x 0 90 180 270 360 450 540 630
720
Y = sin 0 1 0 -1 0 1 0 -1
x 0

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Plotting the curve of y against x

The cosine graph ( y = cos x)

The graph of y = cos x is called the cosine wave
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Draw the cosine graph for 00 to 7200
Solution
We tabulate a table as follows
Substituting the value of x in cos x to get y.
x 0 90 180 270 360 450 540 630 720
Y= 1 0 -1 0 1 0 -1 0 1
cos x

On graph paper
The amplitude is 1 unit
Period = 3600

Example
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 a) Draw the graph of y = 2 sin 2x
b) Find the amplitude and period

Solution
Work out the value of y by substituting the value of x to be between 00
to 3600 at an interval of 450

Formulate a table as below

x 0 45 90 135 180 225 270 315 360
2x 0 90 180 270 360 450 540 630 720
Sin 0 1 0 -1 0 1 0 -1 0
2x
Y= 0 2 0 -2 0 2 0 -2 0
2sin
2x

Plotting the value of y against the x

On graph paper
Amplitude = 2 units
Period = 1800
= 1800
NOTE:
In the graph y = m sin nx or y = m cos nx where m and n are constant.
Amplitude = m
Period = 360 i.e., y = 2 sin 2x
n

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 Amplitude= 2
Period = 360 = 1800
2

Exercise
Determine the amplitude and period in the following graph
a) Y = 3 sin 2x
b) Y = 4 cos 1 x
2

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Some transformation of waves
Draw the graphs of y = sin x and y = 2 sin x on the same axes for 0 0
0
¿ x ≤720

x 00 900 1800 2700 3600 4500 5400 6300 7200

Y= 0 1 0 -1 0 1 0 -1 0
sinx
Y=2 0 2 0 -2 0 2 0 -2 0
sin x

ON GRAPH PAPER
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Amplitude of y = sin x = 1
Y = 2 sin x = 2
Period is same for y = sin x and y = 2 sin x
From the graph the wave y = 2 sin x can be obtained from y= sin x by
applying a stretch of factor 2 with x – axis invariant.

Example
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By drawing the waves y = sin x and y = 3 sin 1 x on the same axes.
2
Describe how to obtain y = 3 sin 1 x from y = sin x
2
State the period and amplitude of y = 3 sin 1 x
2
Solution
Formulating the table

x 0 90 180 270 360 450 540 630 720 810 900

Y= 0 1 0 -1 0 1 0 -1 0 1
sin
x
1
2
x 0 45 90 135 180 225 270 315 360 405 450
Sin 0 0.7 1 0.7 0 -0.7 -1 -0.7 0 0.7 1
1
x
2
Y= 0 2.1 3 2.1 0 -2.1 -3 -2.1 0 2.1 3
3sin
1x
2

Graph on graph paper

Amplitude of y= 3 sin 1 x is 3
2

Period of y = 3 sin 1 x is 7200

2

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Y = 3 sin 1 x can be obtained from y = sin x by a sketch of scale factor 3
2
with x- axis invariant followed by a sketch factor 2 with y – axis invariant
Example
Compare the waves of y = sin x and y = sin (x + 30)

Solution
Formulating the table to get the values of y as follows.

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x 0 3 6 9 1 1 1 2 2 2 3 3 3 3 4 4 48
0 0 0 2 5 8 1 4 7 0 3 6 9 2 5 0
0 0 0 0 0 0 0 0 0 0 0 0
Y = sinx 0 0. 0. 1 0.8 0.5 0 - - -1 - - 0 0.5 0.8 1 0.8
5 87 7 0.5 0.8 0.8 0.5 7 7
7 7
(x+30) 3 60 90 12 15 18 21 24 27 30 33 36 39 42 45 48 51
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
Y=sin(x 0. 0. 1 0. 0.5 0 - - -1 - - 0 0.5 0.8 1 0.8 0.5
+30) 5 87 87 0.5 0.8 0.8 0.5 7 7
7 7

The amplitude and periods of the two waves are the same
Amplitude is 1 unit
Period is 3600
The wave of y = sin x leads y = sin (x + 30) by 300. This angle difference
300 is known as phase angle.

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Y = sin (x + 30) can be obtained from y = sin x by a transaction of ( −300 ) .
If the equation is in the form
Y = k sin (bx ±Ө) or
y = k cos (bx ±Ө)
Where k and b are constants
Amplitude = k
Period = 360
b

and angle Ө is the phase angle

Exercise
1. Draw on the same axes the graphs of
a) y = cos x and y = cos 1 x for 00 ¿ x ¿ 720
2

b)y = sin x and y = sin (x – 30) for 0 ¿ x ¿ 540

In each case describe the transformation that maps the first graph
on to the second
2. Describe fully the transformations the graph of
Y = cos x onto the graph of y = 3 cos x

More exercise
- KLB book 4 page 99 – 100
- Advancing in mathematics book 4 page 62-63

Trigonometric equations
In trigonometric equations there are an infinite number of roots and
therefore we specify the range of values required.
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Example
Solve for x in the equations
a) Cos x = 1 0 ¿ x≤360
2

The x = cos -1
X = 600
1 is positive hence cosine is positive in the first and fourth quadrant
2
hence the solution is x = 600 – 1st Quadrant
X= 360 – 60 4th quadrants
X = 3000

b) 2 cos 2x = 1 00 ¿ x ¿ 360
Dividing both sides by 2
2 cos ⁡2 x = 1
2 2

Cos 2x = 1
2

2x = cos-1 1 cos -1 1 = 60
2 2
Cos is positive in 1st and 4th Quadrant
1st 2nd 1st 4th
2x = 60, 360 – 60, 360 + 60, 360 + 300

2nd rotation
1st rotation

2x = 60, 300, 420, 660 dividing by 2

2x = 60 , 300 , 420 , 660
2 2 2 2 2

X = 300, 1500, 2100, 3300

3) Solve 8 sin2Ө + 2 sin Ө – 3 = 0 for 00 ¿ Ө ¿ 180

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Solution
Sin2Ө can be written as (sin)2
Let sin Ө = y
8(sin Ө)2 + 2 sin Ө – 3 = 0 substituting sin Ө with y
8(y)2 + 2y – 3 = 0 quadratic equation
Solve for y by factorization
M x N = 8 x -3
= -24
M + N =2
The numbers are 6 and -4
-Substituting the coefficient of x with the numbers.
8y2 + 6y – 4y – 3 = 0 -Introducing bracket by pairing
(8y2 + 6y) – (4y -3) = 0 -The negative in 4y – 24 will change to
positive
(8y2 + 6y) – (4y + 3) = 0 -Removing the common outside the
brackets
2y(4y + 3) + (4y + 3) = 0
4y + 3 is common factorize
4y + 3(2y – 1) = 0
(4y + 3) (2y-1) = 0
This means
4y + 3 = 0 or 2y – 1=0
4y = -3 or 2y = 1
Y= −3 or y = 1
4 2

But sin Ө= y substituting value

Sin Ө = −3 or sin Ө = 1
4 2

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Ө = -(sin -1 3 ) or Ө = sin -1 1
4 2

Ө = 48.60 (1 d.p) Ө = 30
Sin Ө = - (48.6)
Since is negative in the 3rd and 4th quadrant
3rd Quadrant = 180 + 48.6
= 228.60
4th Quadrant = 360 – 48.6
= 311.40
For sin Ө = 30
Sin is positive in 1st and 2nd Quadrant so that the value for
Ө = 1st = 30
2nd = 180 – 30
= 150
Values for Ө = 30, 150, 228.6, 311.4
But will rewrite the range o to 1800
Hence Ө = 300 and 1500

Exercise
1. Solve the equation 4 cos 2Ө = 1 00 ¿Ө ¿ 360
0

2. Solve the equation

4 sin 2x + 4 cos x = 5 for 00 ¿ x ≤360
0

3. Solve the equation 2 cos (2Ө) = 1 for 00 ¿ x ¿ 3600

Use of graphs to solve trigonometric equation

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 Graphical method of solving equation drawing two independent graphs
on the same axes and getting the roots.
Example
i. Draw the graph of y = cos x and y = 2 cos (x + 30) for 00 ¿ x
0
¿ 360
ii. Hence solve the equation cos x – 2 cos (x + 30) = 0

Solution
Formulate the table at interval of 300 and work the value of y.
x 0 30 60 90 120 150 180 210 240 270 300 330 360
Y= cos x 1 0.87 0.5 0 -0.5 -0.87 -1 -0.87 -0.5 0 0.5 0.87 1.0
(x+30) 30 60 90 120 150 180 210 240 270 300 330 360 390
Cos(x+30) 0.87 0.5 0 -0.5 -0.87 -1 -0.87 -0.5 0 0.5 0.87 1.0 0.87
Y=2 1.73 1.0 0 -1 -1.73 -2 -1.73 -1 0 1 1.73 2 1.73
cos(x+30)

ON GRAPH PAPER
ii.Cos x – 2cos (x+ 30) = 0
Cos x = 2 cos (x + 30) hence the roots are the point of intersection at
x= 380, 2170

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Exercise
1a. Draw the graph of y = sin (x + 30) and y = cos (x-15)0. For -300 ¿ x ¿
2700 on the same axes.
b) Use your graph to solve
cos (x – 150) – sin (x + 30) = 00

2a) draw the graph y = 4 cos 2x and y = 2 sin (2x + 30) for 00 ¿ x ¿ 1800
b) Use your graph to solve
4 cos 2x – 2 sin (2x + 30) = 0

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More exercise
- KLB book 4 page 102 – 103
- Advancing in mathematics page 66 - 67

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