Trigonometry III - Mathematics Form 4 Notes

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Introduction

  • Consider the right – angled triangle OAB
    right angle triangle OAB
    sinθ = AB
               r
    AB = rsinθ
    OA = rcosθ
    Since triangle OAB is right- angled
    OA2 + AB2 = OB2(pythagoras theorm)
    (rcosθ)2 + (rsinθ)2 = r2
    Divide both sides by r2 gives
    cos2θ + sin2θ = 1
    tanθ = sinθ
               cosθ

    Example
    If tanθ = a show that;
    cosθ(sin2θ+cos2θ) = 1
            sinθ                  a

    Solution
    Factorize the numerator gives and since sin2θ + cos2 θ= 1
    But sinθ = tanθ
          cosθ
    Therefore, = cosθ =  1    = 1
                       sinθ    tanθ     a

    Example
    Show that
    (1-cosθ)(1+cosθ) = tan2θ
    (1-sinθ)(1+sinθ) 
    Removing the brackets from the expression gives
    1 -cos2θ                  reason [(A-B)(A+B)=(A2-B2)]
    1-sin2θ 
    Using sin2θ + cos2θ = 1
    sin2θ + 1 = cos2θ
    Also
    1-cos2θ = sin2θ
    Therefore
    1 -cos2θ =sin2θ =  tan2θ
    1-sin2θ     cos2θ

    Example
    Given that sinθ = 1/3, find
    1. cos2θ
    2. tan2θ
    3. tan2θ + cos2θ

    Solution
    Using the right angle triangle below.
    right angle triangle example
    1.  cosθ = √8
                   3
      therefore cos2θ = (√8/3)2 =8/9
    2. tan2θ = ( 1/√8)2= 1/8
    3. tan2θ + cos2θ = 1/8 + 1/9 =1 1/72


Waves

Amplitude

  • This is the maximum displacement of the wave above or below the x axis.

Period

  • The interval after which the wave repeats itself

Transformations of Waves

  • The graphs of y = sin x and y = 3 sin x can be drawn on the same axis.
  • The table below gives the corresponding values of sin x and 3sinx for 0≤x ≥720o
    x2 0 30 60 90 1 20 1 50 180 210 240 270 300 330 360 390 420 450 480 510 540 570 600 630 660 690 720
    Sin x 0 0.50 0.87 1 .00 0.87 0.50 0 -0.50 -0.87 -0.50 -0.87 -0.50 0 0.50 0.87 1.00 0.87 0.50 0 -0.50 -0.87 -1.00 -0.87 -0.5 0
    3 sinx 0 1 .50 2.61 3.00 2.61 1 .50 0 -1 .50  -2.61 -1 .50  -2.61 -1 .50 0 1.5 2.61 3.00 2.61 1.50 0 -1.50 -2.61 -3.00 -2.61 -2.61 0
  • The wave of y = 3 sin x can be obtained directly from the graph of y = sinx by applying a stretch scale factor 3 , x axis invariant .
    waves introduction
    Note;
  • The amplitude of y= 3sinx is y =3 which is three times that of y = sin x which is y =1 .
  • The period of the both the graphs is the same that is 3600 or 2π

Example

Draw the waves y = cos ½x and y = cos ½x. We obtain y = cos x from the graph y = cos x by applying a stretch of factor 2 with y axis invariant.

waves example
Note;

  • The amplitude of the two waves are the same.
  • The period of y = cos ½x is that is, twice the period of y = cos x


Trigonometric Equations

  • In trigonometric equations, there are an infinite number of roots. We therefore specify the range of values for which the roots of a trigonometric equation are required.

Example

Solve the following trigonometric equations:

  1. Sin 2x = cos x, for 0o ≤ x ≤ 360o
  2. Tan 3x = 2, for 0o ≤ x ≤ 360o
  3. 2 sin(x − π/6)

Solution

  1. Sin 2x = cos x
    Sin 2x = sin (90 – x)
    Therefore 2 x = 90 – x
    X = 30o
    For the given range, x =30o and 150o.
  2. Tan 3x = 2
    From calculator
    3x =63.43o, 243.43o, 423.43o, 603.43o783.43oand 321.14o.

    In the given range;

    x = 21.14o, 81.14o ,141.14o, 201.14o, 261.14o and 321.14o
  3. 2 sin(x − π/6) = − √3
    Sin(x − π/6) = √3/2                 sin (x − π/6) = − √3/2
    x − π/6 = π + π/3, 2π − π/3
    x − π/6 = 4/3π , 5/3π
    x =3/2πc 11/6πc


Past KCSE Questions on the Topic

  1.  
    1. Complete the table for the function y = 2 sin x
      x 0 10o 20 30 40 50 60 70 80o 90 100 110 120o
      Sin 3x 0 0.5000             -0.8660        
      y 0 1 .00             -1 .73        
    2.  
      1. Using the values in the completed table, draw the graph of y = 2 sin 3x for 0o ≤ x ≤ 120o on the grid provided
      2. Hence solve the equation 2 sin 3x = -1.5
  2. Complete the table below by filling in the blank spaces
    X0 00 300 600 90 120 1500 180 2100 2400 270 300 330 3600
    Cos x0 1 .00   0.50     -0.87   -0.87          
    2 cos ½x0 2.00 1 .93       0.52     -1 .00       -2.00
    Using the scale 1 cm to represent 300 on the horizontal axis and 4 cm to represent 1 unit on the vertical axis draw, on the grid provided, the graphs of y = cosx0 and y = 2 cos ½x0 on the same axis.
    1. Find the period and the amplitude of y = 2 cos ½x0
    2. Describe the transformation that maps the graph of y = cos xon the graph of y = 2 cos ½x0
  3.  
    1. Complete the table below for the value of y = 2 sin x + cos x.
      X 00  300  450  600  900  1200  135 150 1800  2250  2700  3150 3600
      2 sinx 0    1 .4 1 .7  2 1 .7 1 .4 1 0   -2 -1 .4 0
      Cos x  1   0.7 0.5  0 -0.5 -0.7 -0.9 -1   0 0.7 1
      Y 1   2.1 2.2 2 1 .2  0.7 0.1 -1   -2 -0.7 1
    2. Using the grid provided draw the graph of y=2sin x + cos x for 0o ≤ x ≤ 360o. Take 1 cm represent 300 on the x - axis and 2 cm to represent 1 unit on the axis.
    3. Use the graph to find the range of x that satisfy the inequalities 2 sinxcosx > 0.5
  4.  
    1. Complete the table below, giving your values correct to 2 decimal places.
      x 0 10 20 30 40 50 60 70
      Tan x 0              
      2x + 300 30 50 70 90 110 130 150 170
      Sin(2x+300) 0.50     1        
    2. On the grid provided, draw the graphs of y = tan x and y = sin ( 2x + 300) for 00 ≤ x 700
      Take scale:
      2 cm for 100 on the x- axis
      4 cm for unit on the y- axis
      Use your graph to solve the equation tan x − sin ( 2x + 300) = 0.
  5.  
    1. Complete the table below, giving your values correct to 2 decimal places
      X0 0 30 60 90 120 150 180
      2sinx0 0 1   2   1  
      1–cosx0     0.5 1      
    2. On the grid provided, using the same scale and axes, draw the graphs of y = sin x0 and y = 1 – cos x0 00 ≤ x ≤ 1800
      Take the scale:
      2 cm for 30
      0 on the x- axis
      2 cm for 1 unit on the y- axis
    3. Use the graph in (b) above to 
      1. Solve equation
        2 sin x
        o + cos x0 = 1
      2. Determine the range of values x for which 2sinxo > 1 – cosxo
  6.  
    1. Given that y = 8 sin 2x – 6 cos x, complete the table below for the missing values of y, correct to 1 decimal place.
      X 00  150  300  450  600  750  900  1050 1200
      Y = 8sin2x – 6cosx  -6  -1 .8     3.8 3.9 2.4 0   -3.9
    2. On the grid provided, below, draw the graph of y = 8 sin 2x – 6 cos for 0≤ x ≤ 1200
      Take the scale
      2 cm for 15
      0 on the x- axis
      2 cm for 2 units on the y – axis
    3. Use the graph to estimate
      1. The maximum value of y
      2. The value of x for which 4 sin 2x – 3 cos x =1
  7. Solve the equation 4 sin (x + 300) = 2 for 00 ≤ x ≤ 3600
  8. Find all the positive angles not greater than 1800 which satisfy the equation
    Sin2x – 2 tan x = 0
    Cos x
  9. Solve for values of x in the range 00 ≤ x ≤ 3600 if 3 cos2x – 7 cos x = 6
  10. Simplify 9 – y2 where y = 3 cos θ
                   y
  11. Find all the values of θ between 00 and 3600 satisfying the equation 5 sin θ = −4
  12. Given that sin (90 – x) = 0.8. Where x is an acute angle, find without using mathematical tables the value of tan x0
  13. Complete the table given below for the functions
    y= −3 cos 2x0 and y = 2 sin (3x/20 + 300) for 00 ≤ x ≤ 1800
    X0 00 200 400 600 800 1000  120 140 1600  1800
    -3cos 2x0 -3.00  -2.30  -0.52  1 .50  2.82 2.82 1 .50 -0.52 -2.30  -3.00
    2sin(3x0 +300) 1 .00  1 .73  2.00 1 .73  1 .00  0.00 -1 .00  -1 .73  -2.00  -1 .73
    Using the graph paper draw the graphs of y = -3 cos 2x0 and y = 2 sin (3x/20 + 300)
    1. On the same axis. Take 2 cm to represent 200 on the x- axis and 2 cm to represent one unit on the y – axis
    2. From your graphs. Find the roots of 3 cos 2 x+ 2 sin (3x/20 + 300) = 0
  14. Solve the values of x in the range 00 ≤ x ≤ 3600 if 3 cos2x – 7cos x = 6
  15. Complete the table below by filling in the blank spaces
    x0 0 30 60 90  120  150 180  210 240  270  300  330  360
    Cosx0  1.00    0.50     -0.87   -0.87          
    2cos½x0  2.00  1 .93          0.5            
    Using the scale 1 cm to represent 300 on the horizontal axis and 4 cm to represent 1 unit on the vertical axis draw on the grid provided, the graphs of y – cos x0 and y = 2 cos ½ x0 on the same axis
    1. Find the period and the amplitude of y =2 cos ½ x0
      Ans. Period = 7200. Amplitude = 2
    2. Describe the transformation that maps the graph of y = cos x0 on the graph of y = 2 cos½ x0
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