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Higher Mathematics Unit 1 Trigonometric Functions and Graphs

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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

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

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1 2 radians = 360° radians = 180°

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

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0°0° 90° 180° 270° 0 2 33 2 22

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

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

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

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radians = = 45° Radians to Degrees: Change radians to degrees 4 radians = 180° 4 4 180° radians = = 270° 33 2 2 3 180° Change radians to degrees 33 2

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We can also convert as follows. Radians Degrees Convert to degrees 180 55 6 Radians 5 180 6 5 180 6 150°

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180 4 4 Radians Change to degrees 2 2 3 Radians 3 3 4 5 5 3 2 180 3 3 180 4 5 180 3 45° 120° 135° 300°

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The angles in the following table must be known. They are essential for non-calculator questions. remember as factors or multiples of 180 ° 360 180 90 60 45 30 Degrees Radians 22 2 3 4 6 radians = 180°

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120° 135° 210° 270° 315° 360° 55 6 55 4 44 3 55 3 Most angles in non-calculator work are multiples of those above Use them to complete the table below Degrees Radians 22 3

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120° 135° 150° 210° 225° 240° 270° 300° 315° 360° 22 3 33 4 55 6 77 6 55 4 44 3 33 2 55 3 77 4 11 6 Degrees Radians

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Sketching Trig Graphs

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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

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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 = 33 2 2 0 22

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y = asinx

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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

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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°

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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 33 2

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y = acosx

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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 33 2

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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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y = -sinx

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

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y = -cosx

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

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y = sin nx

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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

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y = cos nx

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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

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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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Adding or subtracting from a Trig Function

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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.

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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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Adding or subtracting from x

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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.

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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)

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

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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

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

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Ratios and Exact Values Exact Values for 45° 1 1 45° x x² = 1² + 1² x² = 2 x = √ 2 √2

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Ratios and Exact Values Exact Values for 45° 1 1 45° Sin 45° = Cos 45° = Tan 45° = √2 1 1 1

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Ratios and Exact Values Exact Values for /4 1 1 /4 Sin /4 = Cos /4 = Tan /4 = √2 1 1 1

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Ratios and Exact Values Exact Values for 30° & 60° 60° 30° 2 11 60° 30° 1 2 60° 2 2 2 Equilateral Triangle

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

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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

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/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

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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

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/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

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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

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TAN positive ALL Positive COS Positive SIN positive Angles Greater than /2 0 /2 3 /2 22

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22 /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

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ALL SIN TAN COS ° ° 270° 360° x°x° (180 - x)° (180 + x)° (360 - x)°

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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

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

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

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

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Example 3

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ALL SIN TAN COS /2 3 /2 22 ( - ) ( + ) (2 - )

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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

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

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Cos = Example 4 √2 1 1 /4 √2 cos = cos /4 = = /4 √2 1 1

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

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Example 4

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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.

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

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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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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

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sin x = sin 45° = x = 45° Example 6 √2 1 1 1 ° √2 1

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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

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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

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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 = -0.75 Example 7 4 -3

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sin = -0.75 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

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= 0.85 radians = + 0.85or = 2 - 0.85 = 3.14 + 0.85or = 6.28 - 0.85 = 3.99 or 5.43 radians Example 7

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

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Solve cos² x° + sin x° = 1, for 0 ≤ x ≤ 360 (substitute cos² x° = 1 - sin²x° into the equation) 1 - sin² x° + sin x° = 1 1 - 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

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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 = 36.87 or 180 - 36.87 x = 36.87° or 143.13° Example 9

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x = 36.87° or 143.13° The function we are considering is sin (2x - 20) Therefore 2x - 20 = 36.87 or 2x - 20 = 143.13, 2x = 56.87 or 2x = 163.13 x = 28.4°orx = 81.6° Example 9

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

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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 - 1.23 (remember to put calculator in radians) = 1.23 or 6.28 - 1.23 = 1.23 or 5.05 radians Example 10

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= 1.23 or 5.05 radians The function we are considering is cos(2 + /4) 2 + /4 = 1.23or 2 + /4 = 5.05 2 = 1.23 - 0.79or 2 = 5.05 - 0.79 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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