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The Wave Function

The Wave Function. Expressing a cos x + b sin x in the form k cos( x ±  ) or k sin( x ±  ). Heart beat.

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The Wave Function

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  1. The Wave Function Expressing a cos x + b sin x in the form k cos(x ± ) or k sin(x ± ) Heart beat 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 Electrical

  2. General shape for y = sin x + cos x • Like y = sin x shifted left • Like y = cos x shifted right • Vertical height (amplitude) different y = sin x +cos x y = sin x y = cos x

  3. 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: 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 . With these constants the expressions on the right hand sides = those on the left hand side FOR ALL VALUES OF x

  4. Write 4 cos xo + 3 sin xo in the form k cos(x – )o, where 0 ≤  ≤ 360 sin tan  = cos cos(x – ) = cos x cos  + sin x sin  k cos(x – )o = k cos xo cos  + k sin xo sin  4 cos xo 3 sin xo + Now equate with 4 cos xo + 3 sin xo It follows that : k cos  = 4 and k sin  = 3 cos2x + sin2x = 1 (kcos )2 + (ksin )2 = k2 k2= 42 + 32 tan  = ¾ k = √25  = tan-1 0∙75 k = 5  = 36∙9 4 cos xo + 3 sin xo = 5 cos(x – 36∙9)o

  5. Write cos x – √3 sin x in the form R cos(x + ), where 0 ≤  ≤ 2π sin tan  = cos 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 cos2x + sin2x = 1 (Rcos )2 + (Rsin )2 = R2 R2= 12 + (√3)2 tan  = √3/1 R = √4  = tan-1 √3 R = 2  =π/3(60) cos x – √3 sin x = 2 cos(x + π/3)

  6. Write 5 cos 2x + 12 sin x in the form k sin(2x + ), where 0 ≤  ≤ 360 sin tan  = cos sin(2x + ) = sin 2x cos  + cos 2x sin  k sin(2x + ) = k sin 2xo cos  + k cos 2xo sin  5 cos 2xo + 12 sin 2xo It follows that : k cos  = 12 and k sin  = 5 cos2x + sin2x = 1 (kcos )2 + (ksin )2 = k2 k2= 122 + 52 tan  = 5/12 k = √169  = tan-1 0∙417 k = 13  = 22∙6 5 cos 2x + 12 sin 2x = 13 sin(2x + 22∙6)

  7. Maximum and Minimum Values MAX cos x = 1 MAX sin x = 1 when x = 90o when x = 0o or 360o MIN cos x = –1 MIN sin x = –1 when x = 180o when x = 270o MAX k cos (x ± ) is 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 when (x ± ) = 90 (π/2) MIN k sin (x ± ) is – k when (x ± ) = 270 (3π/2)

  8. 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. cos –ve sin +ve A S sin tan  = T C cos k cos(x – )o = k cos xo cos  + k sin xo sin  sin xo – cos xo k cos  = –1 k sin  = 1 k2= (–1)2 + 12 k = √2 tan  = – 1 MAX f(x) = √2 angle = tan-1 1 = 45o When angle = 0  = (180 – 45)o = 135o x – 135 = 0 f(x) = √2 cos (x – 135)o x = 135o

  9. Maximum and Minimum Values A synthesiser adds two sound waves together to make a new sound. The first wave is described by V = 75sin to and the second by V = 100cos to, 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

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

  11. remember K = 25k =25 × 5 = 125 The minimum value of sin is -1 and it occurs where the angle is 270o Therefore, the minimum value of Vresult is – 125 Adding or subtracting 360o leaves the sin unchanged

  12. Minimum, we have:

  13. Solving Trig Equations

  14. 90o A S 180o 0o C T 270o cos is +ve sin is +ve

  15. Re-write the trig. equation using your result from step 1, then solve. 90o A S 180o 0o C T 270o cos is +ve

  16. 90o A S 180o 0o C T 270o cos is –ve sin is –ve

  17. 2x – 213.7 = 16.1o , (180-16.1o),(360+16.1o),(360+180-16.1o) 2x – 213.7 = 16.1o , 163.9o, 376.1o, 523.9o, …. 2x = 229∙8o , 310∙2o, 589∙9o, 670∙2o, …. x = 114∙9o , 188∙8o, 294∙9o, 368∙8o, ….

  18. Part of the graph of y = 2sin x + 5cos x is shown • Express y = 2sin x + 5cos 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): sin + , cos + Equate coefficients: Square and add Dividing: Put together: Minimum when: P has coords.

  19. 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. Expand k sin(x - a): a is in 1st quadrant (sin and cos are +) Equate coefficients: Square and add Dividing: Put together: Sketch Graph

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