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Trigonometric ratios 3 Questions

 

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

 

X0

00

300

600

900

1200

1500

1800

2100

2400

2700

3000

ecolebooks.com

3300

3600

Cos x

1.00

 

0.50

  

-0.87

 

-0.87

     

2cos½x

2.00

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

Image From EcoleBooks.com

15

30

45

60

75

90

3cos2x-1

-40

-3.6

 

-1.0

0.5

1.6

 

1.6

0.5

 

-2.5

-3.6

-4.0

2sin (2x+30)

-1.0

-1.73

 

-1.73

-1.0

0

 

1.73

2.0

 

1.0

0

-1.0

  1. On the grid provided, draw on the same set of axes the graphs of Image From EcoleBooks.comand Image From EcoleBooks.comforImage From EcoleBooks.com. 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 Image From EcoleBooks.com (1mk)
  3. Solve the equation Image From EcoleBooks.com (2mks)

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

 

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

-1.0

 

1.0

  

0

  

-1.0

 

1.0

y = 2cos x3

-1.4

 

-1.4

  

2.0

  

-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.

 (d) From your graph

  (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  

(c) Use your graph to:

(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

 

Image From EcoleBooks.com8.  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

1.0

  

-1.7

-2

-1.0

  

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

 (c) From your graphs:

  (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  

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

4.00

 

2.00

0

-2.00

-3.46

-4.00

-3.46

-2.00

0

2.00

 

4.00

2 sin (2x +30o )

1.00

1.73

2.00

1.73

 

0

-1.00

-1.73

-2.00

-1.73

 

0

1.00

 

 (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    

 (c) From your graph:-

  (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

 

Trigonometric ratios 3 Answers

 

1.

X0

00

300

600

900

1200

1500

1800

2100

2400

2700

3000

3300

Cos x

1.00

0.87

0.50

0

-0.5

-0.87

-1

-0.87

-0.5

0.5

0.7

1

2cos ½ x

2.00

1.93

1.73

1.41

1

0.52

0.00

-0.52

-1

-1.73

-1.93

-2.00

 

Image From EcoleBooks.com

 

 

  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

1.0

 

1.0

0

 

0

 

y = 2cos x

 

-2.0

 

0

1.4

 

1.4

0

 

-2.0

 

 

b)

 

 

 

 

 

 

 

 

Image From EcoleBooks.com

 

 

 

 

 

 

 

 

 

 

 

 

 

(c) -90o or 90o

(d) (i) Highest point 1 unit

  Lowest point – 1.4

 

Image From EcoleBooks.com2.

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

 

Image From EcoleBooks.comx

240

270

300

330

360

Image From EcoleBooks.comImage From EcoleBooks.com 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)

Image From EcoleBooks.com   = 2units

Image From EcoleBooks.com   y = cos (2x – 30)

  =1unit

 

12o, 159oImage From EcoleBooks.comImage From EcoleBooks.comImage From EcoleBooks.comImage From EcoleBooks.comImage From EcoleBooks.comImage From EcoleBooks.comImage From EcoleBooks.comImage From EcoleBooks.comImage From EcoleBooks.comImage From EcoleBooks.comImage From EcoleBooks.comImage From EcoleBooks.comImage From EcoleBooks.comImage From EcoleBooks.comImage From EcoleBooks.comImage From EcoleBooks.comImage From EcoleBooks.com

 

Image From EcoleBooks.com

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

3.  Image From EcoleBooks.comImage From EcoleBooks.comImage From EcoleBooks.comDetermine the

Image From EcoleBooks.comImage From EcoleBooks.comi) Altitude of the frustrum

 Solution

Image From EcoleBooks.comA1C1 = √ 42 + 42 = √32 = 4√2

Image From EcoleBooks.comImage From EcoleBooks.comImage From EcoleBooks.comImage From EcoleBooks.comImage From EcoleBooks.comImage From EcoleBooks.comAC = √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

Image From EcoleBooks.comImage From EcoleBooks.com

Image From EcoleBooks.comImage From EcoleBooks.comImage From EcoleBooks.comii) Angle between AC and the base ABCD

AX = 3√2 + 4√2 = 7√2

Image From EcoleBooks.comImage From EcoleBooks.com Tan ø = CX/AX = √7/ 7√2 = 2.646/9.898

Image From EcoleBooks.comImage From EcoleBooks.comImage From EcoleBooks.com = 0.2673

 = tan -10.2673

= 14.96°

Image From EcoleBooks.com

iii) Volume of pyramid = 1/3 bh

Image From EcoleBooks.comAC = 10 √ 2

Image From EcoleBooks.comImage From EcoleBooks.comA1C1= 4 √ 2

Image From EcoleBooks.comL.S.F = 10:4

Image From EcoleBooks.comImage From EcoleBooks.com
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.  Image From EcoleBooks.com(a) table completed

Image From EcoleBooks.com(b)

Image From EcoleBooks.com(c) (i) 3 P1 – plotting

Image From EcoleBooks.com   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.

 

 

Image From EcoleBooks.com

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

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

-1.0

  

-1.7

 

0

   

 

 

 

 

Image From EcoleBooks.com

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

(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

 

 

 

 

 

 

 

 

 

 

 

 

 

Image From EcoleBooks.com

 
 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

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.    

 

 

 

 

 

Image From EcoleBooks.com

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

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

 

 

Image From EcoleBooks.com(b) graph

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

(c)(i) Amplitude = 4

  (ii) period = 180o

 

 (d) x = 30o, 120o

 

 


 




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