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	-1.0		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 0^o ≤x ≤360^o
      

      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 0^o ≤x ≤360^o
   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
   
   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 – 180^o≤ θ ≤120^o
   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 + 30^o)for
      - 180^o ≤ x ≤ 180^o.
      

      	-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
   
   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 30^o 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)
   4. 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 ≤ 360^o
    

    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°.
    2. 1. State the amplitude of the graph y = cos x^o.
       2. State the period of the graph y = 2 cos (x + 30°).
    3. Use your graph to solve
       Cos x = 2cos(x+30°)
14. Solve the equation sin(2θ+10) = -0.5 for 0 ≤ θ≤ 2π^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 +30^o) for 0^o≤X ≤180^o. Take the scale: 1cm for 15^o 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 + 30^o)
    4. Use your graph to solve:-
       4cos2x – 2sin (2x +30) = 0

Answers

1.   
   1.  
      

      Xo	-225	-180	-135	-90	-45	0	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.  
      
      
   3. -90^o or 90^o
   4. 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

      
      
   3. 1. Amplitudes:, y = 2 sin ( x + 15)
         = 2units
         y = cos (2x – 30)
         =1unit
         12^o, 159^o
         
         
         
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
      
4. 1. table completed
   2.  
      
   3. 1. 3 P1 – plotting
         S1- scale
         C1 – smooth curve
      2. 180^o
      3.  Line y = 1 drawn
         x = 4.5^o or 72.8^o – 107.2^o - 175.4^o
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. √3x ½
   2
     1   x  1 
   √3     √2
   √3 x √6
    4       1
   √18
     4
   3 √2
     4
8.  
   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	-1.0			-1.7		0

   2.  
      
      
   3. 1. Amplitude =3
      2. x = 36^o
         x = 216^o
      3. 33^{o ≤}x ≤213^o
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 = 30^o
    x = 0
    0, 180, 360
12.   
    1.   
    2.   
    3. 10sin x = ^-1/₅₀ + 5
       Y = ^-1/₅₀ + 5
       

       X	0	50
       y	5	4

       X1 = 28^o ±1
       X2 = 70^o ±1
13.   
    1.  
       
       
    2. 1. amplitude = 1
       2. Period = 360°
       3. 45°, 219°
14. 2θ+ 10 = 210^o, 330^o, 570^o, 690^o
    2θ= 200, 320, 560, 680
    = 100^o, 160^o, 280^o, 340^o
    = 5π^c , 8π^c, 14π^c, 17π^c
       90     9      9       9
15.  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°
16.  
    1.  
       

       x	15o	60o	150o	165o
       4cos2x	3.46			3.46
       2sin(2x+30o)		1.00	-1.00

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