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 ±  )

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

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 .

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 = 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 

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 = (√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 

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 = 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 

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

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) 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

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

Expand and equate coefficients C A S T 0o0o 180 o 270 o 90 o cos is +ve sin is –ve

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

Minimum, we have:

Solving Trig Equations

C A S T 0o0o 180 o 270 o 90 o cos is +ve sin is +ve

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

C A S T 0o0o 180 o 270 o 90 o cos is –ve sin is –ve

2x – = 16.1 o, ( o ),( o ),( o ) 2x – = 16.1 o, o, o, 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, ….

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 +

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 +)