Trigonometric Ratios 3 Questions and Answers - Form 4 Topical Mathematics

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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 -1.0   1.0     0     -1.0    1.0
      y=2cos x3 -1.4   -1.4     2.0     -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.
    4. From your graph
      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
    3.  Use your graph to:
      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
    trigonometric ratio q3
    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 
    trigonometric ratio q5
  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
    trigonometric ratio q7
    leaving your answer in surd form.
  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 1.0     -1.7 -2 -1.0     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
    3. From your graphs:
      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
      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 4.00   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 0 -1.73 0 1.00
    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
    3. From your graph:-
      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

Answers

  1.   
    1.  
      Xo  -225  -180  -135  -90  -45  45  90  135  180  225
      y=sin2x   0   0 1.0   1.0 0   0  
      y=2cosx   -2.0   0 1.4   1.4 0   -2.0  
    2.  
      trigonometric ratio answer 1b
    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

        trigonometric ratio ans2
  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
      trigonometric ratio ans3i
    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°
      trigonometric ration ans3ii
    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
      trigonometric ratio ans3iii
    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.  
    Trigonometric ratio ans6
  6. √3x ½
    2
      1   x  
    √3     √2
    √3 x 6
     4       1
    √18
      4
    3 √2
      4
  7.  
    1.  
      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 -1.0     -1.7   0      
    2.  
      trigonometric ratio ans 8b
      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
     trigonometric ratio ans10
  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
      50
      4
      X1 = 28o ±1
      X2 = 70o ±1
  12.   
    1.  
      trigonometric ratio ans13
      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
      trigonometric ratio ans16
      1. Amplitude = 4
      2. period = 180o
    3. x = 30o, 120o

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