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Higher Mathematics Unit 1 Trigonometric Functions and Graphs.

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1 Higher Mathematics Unit 1 Trigonometric Functions and Graphs

2 So far we have always measured our angles in degrees. There is another way to measure angles. It is particularly important in applied mathematics. Angles can be measured in RADIANS

3 Radian Measure Length AB = radius AOB subtends an arc equal to a radius

4 1 2  radians = 360°  radians = 180°

5 So 1 radian =  180° So 1 radian ~ 57° ~ 180° =  radians 90° = radians  2 Every 90° is radians  2

6 0°0° 90° 180° 270° 0  2  33 2 22

7 60° = radians (as 180°  ⅓ = 60°) Degrees to radians Change 60° to radians: 180° =  radians  3

8 We can also convert as follows. Degrees Radians Convert 150° to radians (simplifying fraction: divide by 30) 150   180   5 5  6 Radians

9 Change to Radians: 60° = 120° = 210° = 315° = 60         180  3 Radians 2 2 3 7  6 Radians 7 7 4

10 radians = = 45° Radians to Degrees: Change radians to degrees  4  radians = 180°  ° radians = = 270° 33  180° Change radians to degrees 33 2

11 We can also convert as follows. Radians Degrees Convert to degrees 180   55 6 Radians 5      5  °

12      4 Radians Change to degrees 2 2 3 Radians 3 3 4 5 5 3 2      3      5      45° 120° 135° 300°

13 The angles in the following table must be known. They are essential for non-calculator questions. remember as factors or multiples of 180 ° Degrees Radians  22  2  3  4  6  radians = 180°

14 120° 135° 210° 270° 315° 360° 55 6 55 4 44 3 55 3 Most angles in non-calculator work are multiples of those above Use them to complete the table below Degrees Radians 22 3

15 120° 135° 150° 210° 225° 240° 270° 300° 315° 360° 22 3 33 4 55 6 77 6 55 4 44 3 33 2 55 3 77 4 11  6 Degrees Radians

16 Sketching Trig Graphs

17 Trig Graphs The maximum value for sin x is 1 when x = 90° The minimum value for sin x is -1 when x = 270° sin x = 0 (i.e. cuts the x-axis) at: x = 0°, x = 180°, x = 360

18 Trig Graphs The maximum value for sin x is 1 when x = The minimum value for sin x is -1 when x = sin x = 0 (i.e. cuts the x-axis) at: x = x = x = 33 2  2 0  22

19 y = asinx

20 Trig Graphs When sin x is multiplied by a number, that number gives the maximum and minimum value of the function. Note the function still cuts the x-axis at: x = 0,  & 2 

21 Trig Graphs The maximum value for cos x is 1 when x = 0° & 360° The minimum value for cos x is -1 when x = 180° cos x = 0 (i.e. cuts the x-axis) at: x = 90°, x = 270°

22 Trig Graphs The maximum value for cos x is 1 when x = 0 & 2  The minimum value for cos x is -1 when x =  cos x = 0 (i.e. cuts the x-axis) at: x =  2 33 2

23 y = acosx

24 Trig Graphs When cos x is multiplied by a number, that number gives the maximum and minimum value of the function. Note the function still cuts the x-axis at: x =  2 33 2

25 Trig Graphs Using radians, sketch the following trig graphs: y = 5sinx y = 1.5cosx y = 2cosx y = 100sinx When: 0 ≤ x ≤ 2 

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30 y = -sinx

31 The function y = -sinx is a reflection of y = sinx in the x - axis.

32 y = -cosx

33 The function y = -cosx is a reflection of y = cosx in the x - axis.

34 y = sin nx

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36 Trig Graphs When x is multiplied by a number, that number gives the number of times that the graph “repeats” in 2 . i.e. for y = sin nx: period of graph = 2  n PERIOD

37 y = cos nx

38 Trig Graphs When x is multiplied by a number, that number gives the number of times that the graph “repeats” in 2 . i.e. for y = cos nx: period of graph = 2  n PERIOD

39 Trig Graphs Using radians, sketch the following trig graphs: y = 5sin2x y = 4cos2x y = 6cos3x y = 7sin½x When: 0 ≤ x ≤ 2 

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44 Adding or subtracting from a Trig Function

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48 Trig Graphs When a number is added to a trig function the graph “slides” vertically up by that number. When a number is subtracted from a trig function the graph “slides” vertically down by that number.

49 Trig Graphs Using radians, sketch the following trig graphs: y = 3 + sin2x y = cos3x - 4 y = 3sinx + 2 y = 2cos2x - 1 y = 2 - sinx When: 0 ≤ x ≤ 2 

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55 Adding or subtracting from x

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59 Trig Graphs When a number is added to x the graph “slides” to the left by that number. When a number is subtracted from x the graph “slides” to the right by that number by that number.

60 Example 1 Find the maximum turning point, for 0 ≤ x ≤ , of the graph y = 5sin(x -  /3). Consider the function y = 5sin x Maximum value is 5 When x =  /2 For y = 5sin(x -  /3) Max occurs at  /2 +  /3 = 5  /6 Turning Point: (5  /6,5)

61 Example 2 Write down the equation of the drawn function and the period of the graph. Write the function as y = asin bx + c

62 Example 2 y = asin bx + c b = 3 (3 wavelengths in 2  ) Period of graph 2   3 = 2  /3 Difference between max and min = 12 a = 12  2 = 6 y = 6sin 3x + c(graph then shifts up 2) c = +2 y = 6sin 3x + 2

63 Ratios and Exact Values Exact Values for 45° 1 1 Square ° 1 1

64 Ratios and Exact Values Exact Values for 45° ° x x² = 1² + 1² x² = 2 x = √ 2 √2

65 Ratios and Exact Values Exact Values for 45° ° Sin 45° = Cos 45° = Tan 45° = √

66 Ratios and Exact Values Exact Values for  /4 1 1  /4 Sin  /4 = Cos  /4 = Tan  /4 = √

67 Ratios and Exact Values Exact Values for 30° & 60° 60° 30° ° 30° ° Equilateral Triangle

68 60° 30° 1 2 Ratios and Exact Values Exact Values for 30° & 60° x² = 2² - 1² x² = 3 x = √ 3 √3 x

69 60° 30° 1 2 Ratios and Exact Values Exact Values for 30° √3√3 Sin 30° = Cos 30° = Tan 30° = 1 2 √3√3 2 1 √3

70  /3  /6 1 2 Ratios and Exact Values Exact Values for  /6 √3√3 Sin  /6 = Cos  /6 = Tan  /6 = 1 2 √3√3 2 1 √3

71 60° 30° 1 2 Ratios and Exact Values Exact Values for 60° √3√3 Sin 60° = Cos 60° = Tan 60° = 1 2 √3√3 2 √3

72  /3  /6 1 2 Ratios and Exact Values Exact Values for  /3 √3√3 Sin  /3 = Cos  /3 = Tan  /3 = 1 2 √3√3 2 √3

73 COS Positive SIN positive 1 st Quadrant Sin A = (+)ve Cos A = (+)ve Tan A = (+)ve ALL Positive 4 th Quadrant Angles Greater than 90° 0°0° 90° 180° 270° 2 nd Quadrant 3 rd Quadrant Sin A = (+)ve Cos A = (-)ve Tan A = (-)ve Sin A = (-)ve Cos A = (-)ve Tan A = (+)ve TAN positive Sin A = (-)ve Cos A = (+)ve Tan A = (-)ve

74 TAN positive ALL Positive COS Positive SIN positive Angles Greater than  /2 0  /2  3  /2 22

75 22  /2  3  /2 TAN positive ALL Positive COS Positive SIN positive sin 3p/4 cos 7p/6 tan 7p/4 cos 5.4 radians positive negative positive

76 ALL SIN TAN COS  °  ° 270° 360° x°x° (180 - x)° (180 + x)° (360 - x)°

77 Solve 2sin x° = 1, 0° ≤ x ≤ 360° and illustrate the solution in a sketch of y = sin x 2 sin x° = 1 sin x° = ½ Example 3

78 sin x° = ½ Since sin x° is positive it is in the 1 st and 2 nd quadrants

79 Example 3 sin x° = ½ 60° 30° 1 2 √3√3 sin x° = ½ sin 30° = ½ x = 30°

80 Example 3 sin x° = ½ sin 30° = ½ x = 30° x = 30° or x = 180° - 30° x = 30° or x = 150°

81 Example 3

82 ALL SIN TAN COS  /2  3  /2 22  (  -  ) (  +  ) (2  -  )

83 Solve √2cos  +1 = 0, 0 ≤  ≤ 2  and illustrate the solution in a sketch of y = cos  √2cos  +1 = 0 √2cos  = -1 Cos  = Example 4 √2

84 Cos  = Since cos  is negative it is in the 2 nd and 3 rd quadrants Example 4 √2

85 Cos  = Example 4 √2 1 1  /4 √2 cos  = cos  /4 =  =  /4 √2 1 1

86 Cos  = Example 4 √2 cos  = cos  /4 =  =  /4 So  =  -  /4 or  +  /4 = 3  /4 or 5  /4 √2 1 1

87 Example 4

88 Solve cos 3x° = ½, 0° ≤ x ≤ 360° and illustrate the solution in a sketch of y = cos 3x Consider if the equation was cos x = ½ Example 5 As cos x is positive it must be in the 1 st and 4 th quadrants.

89 cos x = ½ Cos 60° = ½ x = 60° or 360° - 60° x = 60° or 300° Example 5 60° 30° 1 2 √3√3

90 However the function we are using is cos 3x Therefore if x = 60° or 300° for cos x = ½ 3x = 60° or 3x = 300°:x = 20° or 100° the graph repeats itself 3 times in 360° with a wavelength of 120° as the function has a wavelength of 120° x = 20° or 100°or 140° or 220°or 260° or 340° Example 5

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92 Solve 2sin² x° = 1 sin² x = ½ sin x = √½ sin x =  As sin x is positive and negative, x will be in all four quadrants Example 6 √2 1

93 sin x = sin 45° = x = 45° Example 6 √  ° √2 1

94 sin x = x = 45° or 180° - 45° or 180° + 45° or 360° - 45° x = 45° or 135° or 225° or 315° Example 6 √2 1

95 Solve 4sin²  + 11sin  + 6 = 0, correct to 2 decimal places, for 0 ≤  ≤ 2  Factorise the equation Consider the equation as:4x² + 11x + 6 = (4x + 3)(x + 2) = 0 4sin²  + 11sin  + 6 = 0 (4sin  + 3)(sin  + 2) = 0 Example 7

96 4sin²  + 11sin  + 6 = 0 (4sin  + 3)(sin  + 2) = 0 4sin  + 3 = 0 or sin  + 2 = 0 4sin  = -3orsin  = -2 sin  = or sin  = -2 (no solution) Therefore we have to solve sin  = Example

97 sin  = As sin  is negative answer must be in 3 rd and 4 th quadrants sin  = 0.75  = sin - ¹ 0.75 (radians)  = 0.85 radians Example 7

98  = 0.85 radians  =  or  = 2   = or  =  = 3.99 or 5.43 radians Example 7

99 Reminders: sin² x° + cos² x° = 1 sin² x° = 1 - cos² x° cos² x° = 1 - sin² x°

100 Solve cos² x° + sin x° = 1, for 0 ≤ x ≤ 360 (substitute cos² x° = 1 - sin²x° into the equation) 1 - sin² x° + sin x° = sin² x° + sin x° -1 = 0 sin x° - sin² x° = 0 sin x°(1 - sin x°) = 0 sin x° = 0or 1 - sin x° = 0 sin x° = 1 x = 0°or 180° or 360°or x = 90° Example 8

101 Solve sin (2x - 20)° = 0.6, correct to 1 decimal place, for 0 ≤ x ≤ 360 Consider if the equation was sin x = 0.6 x = or x = 36.87° or ° Example 9

102 x = 36.87° or ° The function we are considering is sin (2x - 20) Therefore 2x - 20 = or 2x - 20 = , 2x = or 2x = x = 28.4°orx = 81.6° Example 9

103 The function repeats itself twice in 360° i.e. it has a wavelength of 180° x = 28.4° or x = 81.6° or x = ° or x = ° x = 28.4° or 81.6° or 208.4° or 261.6° Example 9

104 Solve 3cos(2  +  /4) = 1, correct to 1 decimal place, for 0 ≤  ≤  Consider if the equation was 3cos x = 1 cos x = ⅓  = 1.23 or 2  (remember to put calculator in radians)  = 1.23 or  = 1.23 or 5.05 radians Example 10

105  = 1.23 or 5.05 radians The function we are considering is cos(2  +  /4) 2  +  /4 = 1.23or 2  +  /4 =  = or 2  =  = 0.44or 2  = 4.26  = 0.2or  = 2.1 (to 1dp) Do not need to add on a wave length of  as 0 ≤  ≤  Example 10


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