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Published byRichard Jared Booker Modified over 9 years ago
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The Wave Function Heart beat Electrical Many wave shapes, whether occurring as sound, light, water or electrical waves, can be described mathematically as a combination of sine and cosine waves. Spectrum Analysis Expressing a cos x + b sin x in the form k cos(x ± ) or k sin(x ± )
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General shape for y = sin x + cos x 1. Like y = sin x shifted left 2. Like y = cos x shifted right 3. Vertical height (amplitude) different y = sin x y = cos x y = sin x +cos x
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Whenever a function is formed by adding cosine and sine functions the result can be expressed as a related cosine or sine function. In general: With these constants the expressions on the right hand sides = those on the left hand side FOR ALL VALUES OF x a cos x + b sin x = k cos(x ± ) or = k sin(x ± ) Where a, b, k and are constants Given a and b, we can calculate k and .
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Write 4 cos x o + 3 sin x o in the form k cos(x – ) o, where 0 ≤ ≤ 360 cos(x – ) = cos x cos + sin x sin k cos(x – ) o = k cos x o cos + k sin x o sin 4 cos x o Now equate with 4 cos x o + 3 sin x o + 3 sin x o It follows that : k cos = 4 and k sin = 3 cos 2 x + sin 2 x = 1 k 2 = 4 2 + 3 2 k = √25 k = 5 tan = ¾ = tan -1 0∙75 = 36∙9 4 cos x o + 3 sin x o = 5 cos(x – 36∙9) o (kcos ) 2 + (ksin ) 2 = k 2 tan = sin cos
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Write cos x – √3 sin x in the form R cos(x + ), where 0 ≤ ≤ 2π cos(x + ) = cos x cos – sin x sin R cos(x + ) = R cos x cos – R sin x sin cos x– √3 sin x It follows that : R cos = 1 and R sin = √3 cos 2 x + sin 2 x = 1 R 2 = 1 2 + (√3) 2 R = √4 R = 2 tan = √ 3 / 1 = tan -1 √3 = π / 3 (60) cos x – √3 sin x = 2 cos(x + π / 3 ) (Rcos ) 2 + (Rsin ) 2 = R 2 tan = sin cos
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Write 5 cos 2x + 12 sin x in the form k sin(2x + ), where 0 ≤ ≤ 360 sin(2x + ) = sin 2x cos + cos 2x sin k sin(2x + ) = k sin 2x o cos + k cos 2x o sin 12 sin 2x o + 5 cos 2x o It follows that : k cos = 12 and k sin = 5 cos 2 x + sin 2 x = 1 k 2 = 12 2 + 5 2 k = √169 k = 13 tan = 5 / 12 = tan -1 0∙417 = 22∙6 5 cos 2x + 12 sin 2x = 13 sin(2x + 22∙6) (kcos ) 2 + (ksin ) 2 = k 2 tan = sin cos
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Maximum and Minimum Values MAX k cos (x ± ) is k kk k when (x ± ) = 0 or 360 (0 or 2π) MIN k cos (x ± ) is – k when (x ± ) = 180 (π) MAX k sin (x ± ) is k kk k when (x ± ) = 90 ( π / 2 ) MIN k sin (x ± ) is – k kk k when (x ± ) = 270 ( 3π / 2 ) MAX cos x = 1 when x = 0 o or 360 o MAX sin x = 1 when x = 90 o MIN cos x = –1 when x = 180 o MIN sin x = –1 when x = 270 o
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Write f(x) = sin x – cos x in the form k cos (x – ) and find the maximum of f(x) and the value of x at which occurs. k cos(x – ) o = k cos x o cos + k sin x o sin sin x o – cos x o k cos = –1 k sin = 1 k 2 = (–1) 2 + 1 2 k = √2 A S T C cos –ve sin +ve tan = sin cos tan = – 1 = (180 – 45) o angle = tan -1 1 = 45 o = 135 o f(x) = √2 cos (x – 135) o MAX f(x) = √2 When angle = 0 x – 135 = 0 x = 135 o
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A synthesiser adds two sound waves together to make a new sound. The first wave is described by V = 75sin t o and the second by V = 100cos t o, where V is the amplitude in decibels and t is the time in milliseconds. Find the minimum value of the resultant wave and the value of t at which it occurs. For later, remember K = 25k Maximum and Minimum Values
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Expand and equate coefficients C A S T 0o0o 180 o 270 o 90 o cos is +ve sin is –ve
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The minimum value of sin is -1 and it occurs where the angle is 270 o Therefore, the minimum value of V result is – 125 Adding or subtracting 360 o leaves the sin unchanged remember K = 25k =25 × 5 = 125
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Minimum, we have:
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Solving Trig Equations
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C A S T 0o0o 180 o 270 o 90 o cos is +ve sin is +ve
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Re-write the trig. equation using your result from step 1, then solve. C A S T 0o0o 180 o 270 o 90 o cos is +ve
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C A S T 0o0o 180 o 270 o 90 o cos is –ve sin is –ve
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2x – 213.7 = 16.1 o, (180-16.1 o ),(360+16.1 o ),(360+180-16.1 o ) 2x – 213.7 = 16.1 o, 163.9 o, 376.1 o, 523.9 o, …. 2x = 229∙8 o, 310∙2 o, 589∙9 o, 670∙2 o, …. x = 114∙9 o, 188∙8 o, 294∙9 o, 368∙8 o, ….
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Part of the graph of y = 2 sin x + 5 cos x is shown a)Express y = 2 sin x + 5 cos x in the form k sin (x + a) where k > 0 and 0 a 360 b) Find the coordinates of the minimum turning point P. Expand ksin(x + a): Equate coefficients: Square and add Dividing: Put together: Minimum when: P has coords. sin +, cos +
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Expand k sin(x - a): Equate coefficients: Square and add Dividing: Put together: Sketch Graph a)Write sin x - cos x in the form k sin (x - a) stating the values of k and a where k > 0 and 0 a 2 b) Sketch the graph of sin x - cos x for 0 a 2 showing clearly the graph’s maximum and minimum values and where it cuts the x-axis and the y-axis. a is in 1 st quadrant (sin and cos are +)
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