Ch.

3 – Trigonometric Functions

Ex 3.01 – Transformations of Trigonometric functions

b) f(x) = 2 tan x y = tan ax

𝜋
period = 1 a=1
=π

Q1 a) horizontal translation changes phase or phase

shift
b) vertical dilation = amplitude
c) horizontal dilation = period
d) vertical translation = centre
Q2. a) y = 5sinx y = k sin ax
amplitude = 5 a=1 c) y = -cosx y = k cos ax
k=5 amplitude = 1 a=1
2𝜋
period = 𝑎 period =
2𝜋
2𝜋 𝑎
= =
2𝜋
1
1
= 2π = 2π
y = -cos x is the reflection of y= cos x about the
x axis

1
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 Ch. 3 – Trigonometric Functions

d)
e)
Q3. a) y = sin x + 1
y = sinx + 1 y = k sin ax + c
amplitude = 1 a = 1, c = 1 b) y = A sin ax
𝑥
k=1 y = sin 2
2𝜋 Amplitude = 1
period = 𝑎 2𝜋
2𝜋 Period = 1
= 1 2

= 2π = 4π
This is the graph of sin x translated 1 unit
upwards

b) y = tan x – 2
c) f(x) = tan 2x
no amplitude involved
c) f(x) = cos x – 3 𝜋
period = 𝑎
𝜋
Q4. a) y = Acos ax =2
y = cos 4x
Amplitude = 1
2𝜋
Period = 𝑎
2𝜋
= 4
𝜋
= 2

2
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 Ch. 3 – Trigonometric Functions

d) f(x) = tan x/4

no amplitude involved
𝜋
period = 𝑎
𝜋
= 1
4
= 4π

Q5. a) y = cos(x + π)
the π shifts the curve π units to the left

3
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 Ch. 3 – Trigonometric Functions

Ex 3.02 – Combining transformations of trigonometric

functions
y = ksin[a(x + b)] + c
y = kcos[a(x + b)] + c

Q1. a) y = 2sinx – 3
amplitude = 2 a=1
k=2 b=0
c = -3
2𝜋
period =
𝑎
2𝜋
= 1
= 2π

b) y = - tan 2x

𝜋
c) f(x) = cos (𝑥 + 2 ) + 1
amplitude = 1 a=1
𝜋
k=1 b= phase shift
2
c=1 vertical translation
2𝜋
period = 𝑎
2𝜋
= 1
= 2π

4
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 Ch. 3 – Trigonometric Functions

Ex. 3.03 – Trigonometric Equations

Q1. y = 2sin 3x is in the usual form
y = A sin(nx)
Amplitude = A
=2
2𝜋
Period = 𝑛 n=3
2𝜋
= 3
a) Analytical solution, solve the equation thus:
2sin 3x = 1
1
sin 3x = 2
1
3x = sin-1(2)
𝜋 5𝜋
=6 or 6
since sin x is positive
st
in the 1 and 2 quadrant nd b) i) -cos x + 3 = x = 1
𝜋 5𝜋
x= or
6×3 6×3
𝜋 5𝜋
= 18 or 18
2𝜋
Other solutions since the period = 3
are
𝜋 2𝜋 13𝜋
18
+ 3 = 18 ,
5𝜋 2𝜋 17𝜋
18
+ 3 = 18 ,

Graphical solution, we graph y = 2sin 3x (done)

and y = 1

ii) -cos + 3 = 2
method 1: solve algebraically
-cos x = 2 – 3
- cos x = - 1
cos x = 1
x = cos-1(1)
= 0, 2π
Q2. a) y = - cos x + 3 [0,2π] method 2: using the graph
and y = x – 1 we plot y = 2
The general equation for cosine curve
y = k cos[a(x + b)] + c
k = -1, hence amplitude = 1
a = 1, co efficient of x
2𝜋
period = 𝑎
2𝜋
= 1
= 2π

5
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 Ch. 3 – Trigonometric Functions

3x = cos-1(-1/2)
2𝜋 4𝜋 2𝜋 4𝜋
=
3 3
, , ( 3 + 2𝜋), ( 3 + 2𝜋), +
2𝜋 4𝜋
( 3 + 4𝜋), ( 3 + 4𝜋),
2𝜋 4𝜋 7𝜋 10𝜋 13𝜋 16𝜋
= 3, 3, 3, 3 , 3 , 3
2𝜋 4𝜋 7𝜋 10𝜋 14𝜋 16𝜋
x= 9, 9, 9, 9 , 9 , 9

Q5 . a)
b)
c)
𝜋
d) cos(x – 2 ) = 0
Q3. a) 2 sin 2x = 1 𝜋
x – = cos-1(0)
1 2
sin 2x = 𝜋 𝜋
2 =2 or − 2
1
2x = sin-1(2) x=2+2
𝜋 𝜋 𝜋 𝜋
or − 2 + − 2 = 0
= 30° basic angle or (180° - 30°) =π
30
x=
2
= 15°, or (180 °- 30°)/2 Q6. a)
= 15°, 75° b)
add a revolution, 2x = 30° + 360°
= 390° Q7. a)
390
x= b)
2
= 195°
add a revolution to 2nd angle = 150° + 360° Q8. a)
= 510° maximum = 20 m, minimum = 6 m
𝟐𝟎 − 𝟔
b) tan 3x = - 1 hence amplitude = 𝟐
3x = tan-1 (-1) A=7m
= -45° period T = 10 seconds
but tan x is also negative in 2nd quadrant hence 2𝜋
T= 𝑛
x = 180 – 45 2𝜋
10 = 𝑛
= 135°
2𝜋
c) n = 10
d) 𝜋
=5
e)
the graph starts at 6 + 7 = 13 m
Q4. a) tan 2x = √3 k = 13
𝜋
2x = tan-1(√3) basic angle = 3 the sine function is y = A sin nx + k
𝜋 4𝜋 𝜋 4𝜋 𝜋
= 3,, ( + 2𝜋), ( 3 + 2𝜋) = 7 sin 𝑡 + 13
3 3 5
𝜋 4𝜋 7𝜋 10𝜋 b) at the maximum height, y = 20 m
= 3, 3 , 3 , 3
𝜋
𝜋 4𝜋 7𝜋 10𝜋 20 = 7 sin 5 𝑡 + 13
x=6, 6, 6, 6 𝜋
𝜋 2𝜋 7𝜋 5𝜋 7 = 7 sin 5 𝑡
x = , , , 𝜋
6 3 6 3
1 = sin 5 𝑡
b) 2 cos 3x + 1 = 0 𝜋
2 cos 3x = -1 5
𝑡 = sin-1(1)
1 𝜋 5𝜋 9𝜋 13𝜋
cos 3x = -
2
=2, , , 2
2 2

6
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 Ch. 3 – Trigonometric Functions

5 25 45 65
t= , , ,
2 2 2 2
= 2.5, 12.5, 22.5, 32.5 seconds
c)

d)

7
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 Ch. 3 – Trigonometric Functions

Test Yourself 3 period = 2π

centre is y = 0

𝑥 𝜋
b) y = tan 2 y = tan ax, period = 𝑎
1 1
= tan 2 𝑥 a=2
𝜋
asymptote normally when tan 2
𝑥 𝜋
2
= 2
x=π
and since the period is 2π the next asymptote
is at 3π
𝜋
period =
𝑎
𝜋
= 1
2
= 2π

Q1. y = 2cos3x – 7
Amplitude = 2
2𝜋
Period = 𝑛
2𝜋
=
3
centre = -7 ans (D)
Q2. to the left the k is positive, in the angle
y = tan(x + π)
Q3. cos 2x = 1 c) y = sin x - 2
2x = cos-1 (1) amplitude = 1
= 0, 2π, 4π period = 2π
0 2𝜋 4𝜋
x = 2 or 2
or 2
centre is y = -2
= 0 or π or 2π
Q4. a) y = 3 cos x
amplitude = 3

8
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 Ch. 3 – Trigonometric Functions

2𝜋𝑡
= - 2π sin( )
3
=0 h’(x) = 0 for min or max
2𝜋𝑡
- 2π sin( 3 ) = 0 ÷ -2π
2𝜋𝑡
sin( 3 ) = 0 shift sin gives
2𝜋𝑡
3
= 0, 𝜋, 2𝜋, 3𝜋, 4𝜋..
2𝜋𝑡
3
=0
t=0
2𝜋𝑡
d) y = -sin x = 3𝜋
3
amplitude = 1 t=3h
period = 2π continuing this t = 9, 12, 15…
centre is y = 0 also for π, 3π, 5π
2𝜋𝑡
=𝜋
3
t = 1.5 h
hence 3π, 5π, t = 4.5, 7.5, 10.5,.
2𝜋𝑡 2𝜋
h” = - 2π cos( )×
3 3
2𝜋 2𝜋𝑡
= ×- 2π cos( )
3 3
−4𝜋2 2𝜋𝑡
= cos( )
3 3
e) y = 3cos2x – 1 −4𝜋2 2𝜋 × 0
t = 0, h”= cos( )
Amplitude = 3 3 3
−4𝜋 2
2𝜋 = ×1
Period = 𝑛 3
2𝜋 −4𝜋2
= = hence max
2 3
−4𝜋 2 2𝜋 × 1.5
=π t = 1.5, h”= cos( )
3 3
centre = -1 −4𝜋 2 2𝜋 × 1.5
= cos( )
3 3
−4𝜋 2
= cos(π)
3
−4𝜋 2
= × -1
3
4𝜋2
= 3 hence min
2𝜋𝑡
The graph of h = 3cos( 3 ) + 10 is shown below. Note
the maximums at A,C,E,G, etc and minimums at
B,D,F, etc. The red line is y = 10 – the equilibrium line
and the amplitude is 3 m. Hence the range of this
graph is 7 ≤ h ≤ 13 which is clearly shown on the
graph. The graph is in terms of y and x rather than h
and t. The dotted lines at y = 7 and 13 illustrate the
tangent needed for the minimum and maximum
2𝜋𝑡 respectively.
Q5. a) h = 3cos( 3
)+ 10 differentiate wrt to t
2𝜋𝑡 2𝜋
h’= - 3 sin( 3 ) × 3
+ 0
2𝜋 2𝜋𝑡
= 3
×-3 sin( 3 )

9
……….
 Ch. 3 – Trigonometric Functions

x = 180 + 30
𝜋 7𝜋
= 210° note; 210° × 180° = 6
𝑟𝑎𝑑
7𝜋
= 6
b) 2sin 2x = 1 [0,2π]
1
sin 2x =2
1
2x = sin-1(2)
𝜋
2x = 6 basic angle
𝜋 5𝜋 𝜋 5𝜋
= 6, , + 2𝜋, 6
6 6
+ 2𝜋
𝜋 5𝜋 13𝜋 17𝜋
x= , , ,
12 12 12 12
2 2
Q6. a) 2cot x + 2 = 2(cot x + 1) but since the angle is 2x the domain is enlarged to
= 2 cosec2 x [0,4π]. Since sin x is positive in the 1st and 2
note: cosec2 x = cot2 x + 1 quadrant then the solutions are these plus 2π as
sin 𝐴 1
b) tan A cosec A = × shown above.
cos 𝐴 sin 𝐴
1
= cos 𝐴 𝜋
c) cos(x – 2 ) = -1 [0,2π]
= sec A 𝜋
c) (sec A + tan A)(sec A – tan A) = sec2 A – tan2A x – = cos-1(-1)
2
𝜋
= 1 + tan2A – tan2A x– =π
2
=1 𝜋
x=π+2
d) sin (180° - x) = sin x 3𝜋
= 2
Q7. a) 4 cos2 x = 3 divide both sides by 4 d)
4 cos2 𝑥 3
=
4 4 Q8. a) maximum = 135, minimum = 85
2 3
cos x=4 take square root of both sides 𝟏𝟑𝟓 − 𝟖𝟓
hence amplitude = 𝟐
3
cos x = ±√4 A = 25
heartbeat = 70 beats/min
√3
=± 2 period = time for 1 beat
√3 1
x = cos-1( 2 ) = 70 𝑚𝑖𝑛
2𝜋
= 30° basic angle T= 𝑛
𝜋 𝜋 𝜋 2𝜋
=6 note; 30° × 180° = 6
𝑟𝑎𝑑 70 = 𝑛
But cos x is positive in the 1st and 4th quadrant. n
𝜋
= 35
𝜋 11𝜋
x = 330° note; 330° × = 𝑟𝑎𝑑 the graph starts at 85 + 25 = 110
180° 6
11𝜋
= 𝑟𝑎𝑑 the sine function is y = A sin nx + k
6 𝜋
Now for the negative case: = 25 sin 𝑡 + 110
35
-1 −√3 b)
x = cos ( )
2
𝜋 5𝜋
= 150° note; 150° × 180° = 6
𝑟𝑎𝑑
5𝜋
= 6
𝑟𝑎𝑑
But cos x is also negative in the 3rd quadrant
hence,

10
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 Ch. 3 – Trigonometric Functions

7𝜋 25𝜋
x= ,
6 4
These are out of the range, we consider in the
negative direction, thus
𝜋 𝜋 𝜋
the angles - 6 𝑎𝑛𝑑 − 6 − 6 .
𝜋 7𝜋
hence we - 6 𝑎𝑛𝑑 − 6
.
𝜋 −𝜋 𝜋 −7𝜋
x- = or x - =
4 6 4 6

−𝜋 𝜋 −7𝜋 𝜋
x= 6
+ 4 x= 6
+4
𝜋 −11𝜋
= and
12 12

Q9. Q10. a) maximum = 80, minimum = 50

𝟖𝟎 − 𝟓𝟎
hence amplitude = 𝟐
A = 15 m
period T = 13 hours
2𝜋
T= 𝑛
2𝜋
13 =
a) 2cos 2x = 1 𝑛
1 2𝜋
cos 2x = n=
2 13
1 the graph starts at 50 + 15 = 65
2x = cos-1(2)
𝜋 k = 65
=3 basic angle the sine function is y = A sin nx + k
but cos x is positive in the 1st and 4th quadrants 2𝜋
= 15 sin 13 𝑡 + 65
𝜋 𝜋
2x = 3 , (2π – 3 )
b) At half way y = 65 m, solve for this condition
𝜋 5𝜋 2𝜋
2x =3, 3 65 = 15 sin 13 𝑡 + 65
Also as the angle is 2x we consider another 2𝜋
0 = 15 sin 13 𝑡
revolution in the opposite direction. Hence we 2𝜋
subtract 2π to the above 15 sin 13 𝑡 = 0 divide by 15
𝜋 5𝜋 𝜋 5𝜋 2𝜋
2x = 3 , , ( − 2𝜋), ( 3 − 2𝜋) sin 13 𝑡 = 0
3 3
𝜋 5𝜋 −5𝜋 −𝜋 2𝜋
= 3, 3 , 3 , 3 𝑡 = sin-1(0)
13
𝜋 5𝜋 −5𝜋 −𝜋 2𝜋
x= , , , 13
𝑡 = 0, π, 2π, ….
6 6 6 6
𝜋 1 13 39
b) tan(x - 4 ) = − 3 t = 0, , 13, , … hours
√ 2 2
𝜋 −1
x - 4 = tan ( − 3)
1 = 0, 6.5, 13, 19.5,…. hours
√
𝜋
=6 Q11. a) the general formula is y = kcos (ax + b)
since tan x is negative in the 2nd and 4th k = amplitude,
quadrants, the required angles are a = constant which determines the period
𝜋 𝜋 5𝜋 11𝜋
(π - 6 ) 𝑎𝑛𝑑 (2π − 6 ) = 𝑎𝑛𝑑 b = phase shift (= 0 in this case)
6 6
𝜋 5𝜋 𝜋 11𝜋 y = cos x given period = π
hence x- = or x - =
4 6 4 6 amplitude = 1
2𝜋
5𝜋 𝜋 11𝜋 𝜋 period =
𝑎
x= 6
+ 4
x= 6
+ 4

11
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 Ch. 3 – Trigonometric Functions

2𝜋
π=
𝑎
aπ = 2π cancel π
a=2
hence the required equation is y = cos 2x
b) amplitude = 5
y = 5 cosx
c) a reflection in the x axis would simply reverse
the sign thus y = - cosx
see graph below

𝜋
d) a phase of 6 units to the right
the general formula is y = kcos (ax + b)
𝜋
b = phase shift (= in this case)
6

𝜋
y = cos(x - )
6
see graph below

e) centre 4 means the grapgh is shifted 4 units up

Hence y = cos x + 4
0r y = 4 + cos x

12
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 Ch. 3 – Trigonometric Functions

Challenge Exercise
Q1.

13
……….