## Trigonometric ratios 3 Questions

1.  Complete the table below by filling in the blank spaces.

 X0 0 300 600 900 1200 1500 1800 2100 2400 2700 3000 ecolebooks.com 3300 3600 Cos x 1 0.50 -0.87 -0.87 2cos½x 2 1.93 0.50

(2mks)

On the grid provided, using a scale of 1 cm to represent 300 on the horizontal axis and 4cm to represent 1 unit on the vertical axis draw the graph of y = cos x0 and y = 2 cos ½ x0.  (4mks)

(a) State the period and amplitude of y = 2 cos ½ x0 (2mks)

(b) Use your graph to solve the equation 2 cos ½ x – cos x = 0.  (2mks)

2.  a) Complete the table below by filling in the blank spaces

 x -90 -75 -60 -45 -30 -15 15 30 45 60 75 90 3cos2x-1 -40 -3.6 -1 0.5 1.6 1.6 0.5 -2.5 -3.6 -4 2sin (2x+30) -1 -1.73 -1.73 -1 0 1.73 2 1 0 -1
1. On the grid provided, draw on the same set of axes the graphs of and for. Using a scale o 1 cm for 150 on axis and 2 cm for I unit on the y-axis   (5mks)
2. State the period of (1mk)
3. Solve the equation (2mks)

3.  Complete the table below by filling in the blank spaces.

 x° 0° 30° 60° 90° 120° 150° 180° 210° 240° 270° 300° 330° 360° Cos x 1.00 0.5 -0.87° -0.87° 2 cos ½ x° 2.00 1.93 0.52 -1.00 -2.00

Using the scale 1 cm to represent 30o on the horizontal axis and 4cm to represent 1 unit on the vertical axis, draw on the grid provided, the graph of y = cos x° and y = 2 cos ½ x°

a)  Find the period and amplitude of y = 2cos ½ x° (2mks)

b)  Describe the transformation that maps the graph of y = Cos x° on the graph of

y = 2 cos ½ x°.  (2mks)

4.  The table below gives some values of y = sin 2x and y = 2 cox is the range given.

(a) 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

(b) On the same axes, draw the graphs of y = sin 2x and y = 2 cos x.

(c) Use your graph to find in values of x for which sin 2x – 2 cos x = 0.

(i) Find the highest point of graph y = sin 2x.

(ii) The lowest point of graph y = 2 cos x.

5.  (a) Copy and complete the table below for y =2sin (x +15)o and y =cos(2x -30)o for 0o
 x  360o

 x 0 30 60 90 120 150 180 210 240 270 300 y =2sin(x+15) y= cos(2x-30)

(b) On the same axis draw the graphs:

y = 2sin (x + 15) and y = cos(2x -30) for  0o
 x  360o

(i) State the amplitudes of the functions y = 2sin (x +15) and y= cos (2x -30)

(ii) Solve the equation 2sin (x+15) – cos (2x -30) = 0

6.  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:

i) Altitude of the frustrum

ii) Angle between AC1 and the base ABCD

iii) Calculate the volume of the frustrum

7.  (a) 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 0.8 21

(b) Use the table to draw the curve y = 3sin (2x +15) for the values – 180o

120o

(c) Use the graph to find:

(i) The amplitude

(ii) The period

(iii) The solution to the equation:-

Sin (2x + 15)o = 1/3

8.  Make q the subject of the formula in A

B

9.  a) 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

b) On the same axis, draw the graphs of y = cos (2x + 45)o and y = -sin (x + 30)o

c) Use the graphs drawn in (b) above to solve the equation.

Cos (2x + 45)o + sin(x + 30)o = 0

10.  Without using tables or calculators evaluate sin 60o cos 60o leaving your answer in surd form. tan 30o sin 45o

11.  (a) 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 3 0 -1.5 -2.6 -3 -1.5 2cosx 1.7 1 -1.7 -2 -1 1.0 1.7 2

(b) 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

(i) State the amplitude of y = 3sin x

(ii) Find the values of x for which 3sin x – 2cos x = 0

(iii) Find the range of values of x for which 3sin x  2cos x

12.  (a) Fill in the following table of the given function:-

 x 0 90 180 270 360 450 540 630 720 810 sin ½x 0 0.71 0 3Sin (½x + 60) -2.6 2.6

(b) On the grid provided draw the graph of the function y = sin ½x
and y = 3Sin (½x + 60)

on the same set of axes

(c) What transformation would map the function y = sin ½
x
onto y = 3 Sin (½ x + 60)

(d) (i) State the period and amplitude of function : y = 3 Sin (½x + 60

(ii) Use your graph to solve the equation: 3Sin ( ½x + 60) – Sin ½x = 0

13.  a) 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 0.50 1 2

b) 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

c) 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

14.  Solve the equation 2 sin (x + 30) = 1 for 0 ≤ x ≤ 360.

15.  (a) 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 10 sin x 0 – 3.4 5.0 7.7 9.4 9.8 10 9.8 9.4 7.7 5.0 3.4 0

(b) 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

(c) By drawing a suitable straight line on the same axis, solve the equation: –

500 sin x = -x + 250

16.  Complete the table below for the functions y = cosx and y =2 cos (x 300) for ≤x ≤ 3600

 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 2 cos (x + 30o) 1.73 0 -1.0 -2.0 -1.73 -1.0 1 1.73 2.00 1.73

(a) On the same axis, draw the graphs of y cos x and y 2cos(x – 30) for O

(b) (i) State the amplitude of the graph y = cos xo.

(ii) State the period of the graph y = 2 cos (x + 30°).

c) Use your graph to solve

Cos x = 2cos(x+30°)

17.  Solve the equation sin(2 +10) = -0.5

for 0

 2c

18.  Solve the equation

4 sin 2x = 5 – 4 cos2
x for 0° ≤ x ≤ 360°

19.  (a) 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 2.00 4 2 sin (2x +30o ) 1 1.73 2 1.73 0 -1 -1.73 -2 -1.73 0 1

(b) On the grid provided; draw on the same axes, the graphs of y = 4cos 2x and

y =2sin(2x +30o)
for 0o X
180o
. Take the scale: 1cm for 15o on the x-axis and

2cm for 1unit on the y-axis

(i) State the amplitude of y = cos 2x

(ii) Find the period of y = 2sin (2x + 30o)

(d) Use your graph to solve:-

4cos2x – 2sin (2x +30) = 0

1.

 X0 0 300 600 900 1200 1500 1800 2100 2400 2700 3000 3300 Cos x 1 0.87 0.5 0 -0.5 -0.87 -1 -0.87 -0.5 0.5 0.7 1 2cos ½ x 2 1.93 1.73 1.41 1 0.52 0 -0.52 -1 -1.73 -1.93 -2

1. amplititude = 2 B1

period = 7200 B1

1. 2cos ½ x = cos x

X = 2220
+60

B1

B1

S1

P1

P1

C1

B1

B1

All values of cos x

All values of cos ½ x

Given scale used

Plotting cos x

Ploting 2 cos ½ x

Curve smooth continous

1.  a)

 Xo -225 -180 -135 -90 -45 0 45 90 135 180 225 y = sin 2x 0 0 1 1 0 0 y = 2cos x -2 0 1.4 1.4 0 -2

b)

(c) -90o or 90o

(d) (i) Highest point 1 unit

Lowest point – 1.4

2.

 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

 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

(i) Amplitudes:, y = 2 sin ( x + 15)

= 2units

y = cos (2x – 30)

=1unit

12o, 159o

3.  Determine the

i) Altitude of the frustrum

Solution

A1C1 = √ 42 + 42 = √32 = 4√2

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

ii) 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

 = tan -10.2673

= 14.96°

iii) 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)

= 441.0/328.224/3

= 413.776/3

= 137.592cm3

4.  (a) table completed

(b)

(c) (i) 3 P1 – plotting

S1- scale

C1 – smooth curve

(ii) 180o

(iii) Line y = 1 drawn

x = 4.5o or 72.8o – 107.2o – 175.4o

5.  (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

6.

7. 7.  3 x ½

2

1 x 1

3 2

3 x 6

4 1

18

4

3
2

4

8. a)

 x 0 30 60 90 120 150 180 210 240 270 300 330 360 3sinx 1.5 2.6 1.5 -2.6 0 2cos x 2 0 -1 -1.7 0

(c) (i) Amplitude =3

(ii) x = 36o

x = 216o

(iii) 33o
 x  213o

9.

 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

10.

 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

11.  Sin (x + 30) = 0.5

x + 30 = 30o

x = 0

0, 180, 360

12.   (c)10sin x = -1/50 + 5

Y = -1/50 + 5

 X 0 50 y 5 4

X1 = 28o
1

X2 = 70o
1

12.

b)  i) amplitude = 1

ii) Period = 360°

iii) 45°, 219°

13.  2 + 10 = 210o, 330o, 570o, 690o

2 = 200, 320, 560, 680

= 100o, 160o, 280o, 340o

= 5c
, 8c, 14c ,17c

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.

 x 15o 60o 150o 165o 4 Cos 2x 3.46 3.46 2Sin (2x + 30o) 1.00 -1.00

(b) graph

(c)(i) Amplitude = 4

(ii) period = 180o

(d) x = 30o, 120o

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