# Trigonometric Ratios 3 Questions and Answers - Form 4 Topical Mathematics

## Questions

1. The table below gives some values of y = sin 2x and y = 2 cox is the range given.
1. Complete
 xo -225 -180 -135 -90 -45 0 45 90 135 180 225 y - sin 2x3 -1 1 0 -1 1 y=2cos x3 -1.4 -1.4 2 -1.4 -1.4
2. On the same axes, draw the graphs of y = sin 2x and y = 2 cos x.
3. Use your graph to find in values of x for which sin 2x – 2 cos x = 0.
1. Find the highest point of graph y = sin 2x.
2. The lowest point of graph y = 2 cos x.
2.
1. Copy and complete the table below for y =2sin (x +15)o and y =cos(2x -30)o for 0ox ≤360o
 x 0 30 60 90 120 150 180 210 240 270 300 y=2sin(x+15) y=cos(2x-30)
2. On the same axis draw the graphs:
y = 2sin (x + 15) and y = cos(2x -30) for 0o x ≤360o
1. State the amplitudes of the functions y = 2sin (x +15) and y= cos (2x - 30)
2. Solve the equation 2sin (x+15) – cos (2x - 30) = 0
3. The diagram below shows a frustum of a square based pyramid. The base ABCD is a square of side 10cm. The top A1B1C1D1 is a square of side 4cm and each of the slant edges of the frustum is 5cm

Determine the:
1. Altitude of the frustrum
2. Angle between AC1 and the base ABCD
3. Calculate the volume of the frustrum
4.
1. Compete the table below:
y = 3sin (2x + 15)o
 x -180 -150 -120 -90 -60 -30 0 30 60 90 120 y 0.8 -0.8 21
2. Use the table to draw the curve y = 3sin (2x +15) for the values – 180o θ ≤120o
3. Use the graph to find:
1. The amplitude
2. The period
3. The solution to the equation:-
Sin (2x + 15)o = 1/3
5. Make q the subject of the formula in

6.
1. Complete the table below for the functions y = cos (2x + 45)o and y = -sin (x + 30o)for
- 180o x 180o.
 -180 -150 -120 -90 -60 -30 0 30 60 90 120 150 180 y=cos(2x+45o) 0.71 -0.97 -0.71 0.71 -0.97 0.97 y=-sin(x+30o) 0.5 0.87 0.5 -0.87 -0.87 0.5
2. On the same axis, draw the graphs of y = cos (2x + 45)o and y = -sin (x + 30)o
3. Use the graphs drawn in (b) above to solve the equation.
Cos (2x + 45)o + sin(x + 30)o = 0
7. Without using tables or calculators evaluate

8.
1. Complete the table below for the functions y = 3 sin x and y = 2 cos x
 X 0 30 60 90 120 150 180 210 240 270 300 330 360 3sin x 2.6 0 -1.5 -2.6 -3 -1.5 2cos x 1.7 1 -1.7 -2 -1 1.0 1.7 2
2. Using a scale of 2cm to represent 1 unit on the y- axis and 1cm to present 30o on the x-axis ,draw the graphs of y =3sinx and y = 2cosx on the same axes on the grid provided
1. State the amplitude of y = 3sin x
2. Find the values of x for which 3sin x – 2cos x = 0
3. Find the range of values of x for which 3sin x ≥2cos x
9.
1. Fill in the following table of the given function:-
 x 0 90 180 270 360 450 540 630 720 810 sin½x 0 0.71 0 3sin(½x+60) -2.6 2.6
2. On the grid provided draw the graph of the function y = sin ½x and y = 3Sin(½x + 60) on the same set of axes
3. What transformation would map the function y = sin ½ x onto y = 3 Sin (½ x + 60)
1. State the period and amplitude of function : y = 3 Sin (½x + 60
2. Use your graph to solve the equation: 3Sin ( ½x + 60) – Sin ½x = 0
10.
1. Complete the table below giving your answer to 2 decimal places
 xº 0o 30o 60o 90o 120o 150o 180o 2sinxo 0 1 2 1 - cos xo 0.5 1 2
2. On the grid provided, using the same axis and scale draw the graphs of :-
y = 2sinxº, and y =1-cosx for
0º≤ x ≤ 180º , take the scale of
2cm for 30º on the x-axis
2cm for 1 unit on the y-axis
3. use the graph in (b)above too solve the equation 2sinx + cosxº = 1 and determine the range of values of for which 2 sinxº =1-cosxº
11.  Solve the equation 2 sin (x + 30) = 1 for 0 x 360.
12.
1. Complete the table below, giving your values correct to 1 decimal place
 x 0o 10o 20o 30o 40o 50o 60o 70o 80o 90o 100o 110o 120o 130o 140o 150o 160o 170o 180o 10sinx 0 3.4 5.0 7.7 9.4 9.8 10 9.8 9.4 7.7 5.0 3.4 0
2. Draw a graph of y = 10 sin x for values of x from 0o to 180o.
Take the scale 2cm represents
20o on the x-axis and 1cm represents 1 unit on the y axis
3. By drawing a suitable straight line on the same axis, solve the equation: -
500 sin x = -x + 250
13. Complete the table below for the functions y = cosx and y =2 cos (x 300) for θ≤x ≤ 360o
 x 0o 30o 60o 90o 120o 150o 180o 210o 240o 270o 300o 330o 360o Cos x 1 0.87 0.5 -0.5 -0.87 -1.0 0.5 0 0.87 1 2cos 1.73 0 -1.0 -2.0 -1.73 -1.0 1 1.73 2.00 1.73
1. On the same axis, draw the graphs of y cos x and y 2cos(x - 30) for O<x < 360°.
1. State the amplitude of the graph y = cos xo.
2. State the period of the graph y = 2 cos (x + 30°).
2. Use your graph to solve
Cos x = 2cos(x+30°)
14. Solve the equation sin(2θ+10) = -0.5 for 0 ≤ θ≤ c
15. Solve the equation
4 sin 2x = 5 – 4 cos2x for 0° ≤ x ≤ 360°
16.
1. Complete the table given below by filling in the blank spaces
 X 0 15 30 45 60 75 90 105 120 135 150 165 870 4cos 2x 4 2 0 -2.00 -3.46 -4 -3.46 -2 0 2 4 2 sin(2x+30o) 1 1.73 2 1.73 0 -1 -1.73 -2 0 -1.73 0 1
2. On the grid provided; draw on the same axes, the graphs of y = 4cos 2x and y =2sin(2x +30o) for 0oX ≤180o. Take the scale: 1cm for 15o on the x-axis and 2cm for 1unit on the y-axis
1. State the amplitude of y = cos 2x
2. Find the period of y = 2sin (2x + 30o)
4. Use your graph to solve:-
4cos2x – 2sin (2x +30) = 0

1.
1.
 Xo -225 -180 -135 -90 -45 0 45 90 135 180 225 y=sin2x 0 0 1 1 0 0 y=2cosx -2 0 1.4 1.4 0 -2
2.

3. -90o or 90o
1. Highest point 1 unit
2. Lowest point - 1.4
2.
1.
 x 0 30 60 90 120 150 180 210 2sin(x+15o) 0.52 1.41 1.93 1.93 1.41 0.52 -0.52 -1.41 Cos(2x -30o) 0.87 0.87 0 -0.87 0.87 0 0.87 0.87
2.
 x 240 270 300 330 360 2sin(x+15o) -1.93 -1.93 -1.41 -0.52 0.52 Cos (2x -30o) 0 -0.87 -0.87 0 0.87

1. Amplitudes:, y = 2 sin ( x + 15)
= 2units
y = cos (2x – 30)
=1unit
12o, 159o

3. Determine the
1. Altitude of the frustrum
Solution
A1C1 = √ 42 + 42 = √32
AC = √102 + 102
= √200
= 10√2
AM + XM = 10√2 - 4√2
= 6√2
AM = 6√2/2 = 3√2
Height = AM =√52 – (3√2)2 = √25 – 18
= √7 = 2.646
the altitude of the frustrum = 2.646 cm

2. Angle between AC and the base ABCD
AX = 3√2 + 4√2 = 7√2
Tan ø = CX/AX =√7/7√2
2.646/9.898
= 0.2673
θ = tan-1 0.2673
= 14.96°
3. Volume of pyramid = 1/3 bh
AC = 10√2
A1C1= 4√2
L.S.F = 10:4
h + 2.646 = 10
h            4
4(h + 2.646) = 10h
4h + 10.584 = 10h
6h = 10.584
h = 1.764
H = h + 2.646
= 1.764 + 2.646 = 4.410
Vf = (1/3 x 10 x 10 x 4.41) – (1/3 x 4 x 4 x 1.76)
=441.0/3 - 28.224/3
=413.776/3
= 137.592cm3
1. table completed
2.
1. 3 P1 – plotting
S1- scale
C1 – smooth curve
2. 180o
3.  Line y = 1 drawn
x = 4.5o or 72.8o107.2o - 175.4o
4.  (A/B)2 = p + 33q
q – 3P
A2q – 3A2P = BP + 3Bq
Aq2 – 3Bq = BP + 3A2P
2(A2 – 3B) = BP + 3A2P
Q = BP + 3A2P
A2 – 3B
5.
6. √3x ½
2
1   x
√3     √2
√3 x 6
4       1
√18
4
3 √2
4
7.
1.
 x 0 30 60 90 120 150 180 210 240 270 300 330 360 3sinx 1.5 2.6 1.5 -2.6 0 2cosx 2 0 -1 -1.7 0
2.

1. Amplitude =3
2. x = 36o
x = 216o
3. 33o ≤x ≤213o
8.
 x 0 90 180 270 360 450 540 630 720 810 sin ½x 0 0.71 1 0.71 0 -0.71 -1 -0.71 0 0.71 3Sin (½x + 60) 2.6 2.9 1.5 -0.78 -2.6 2.9 -1.5 0.78 2.6 2.9
9.
 x 0o 30o 60o 90o 120o 150o 180o 2 sin x 0 1 1.73 2 1.73 1.00 0 1-Cos x 1 0.13 0.50 1 0.06 1.87 2

10. Sin (x + 30) = 0.5
x + 30 = 30o
x = 0
0, 180, 360
11.
1.
2.
3. 10sin x = -1/50 + 5
Y = -1/50 + 5
 X 0 50 y 5 4
X1 = 28o ±1
X2 = 70o ±1
12.
1.

1. amplitude = 1
2. Period = 360°
3. 45°, 219°
13. + 10 = 210o, 330o, 570o, 690o
= 200, 320, 560, 680
= 100o, 160o, 280o, 340o
= c , c, 14πc, 17πc
90     9      9       9
14.  4sin  2x+4cos x-5 = 0
4(1-cos2x) + 4 cosx - 5= 0
4cos2x – 4 cosx + 1=0
4cos2x – 2cosx – 2cos x +1 =0
(2cos x – 1)2 = 0
X = 60°, 300°
15.
1.
 x 15o 60o 150o 165o 4cos2x 3.46 3.46 2sin(2x+30o) 1.00 -1.00
2. graph

1. Amplitude = 4
2. period = 180o
3. x = 30o, 120o

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