2. HIGHER MATHEMATICS Unit 3 - Outcome 4 The Wave Function

3. This topic concerns itself with combining sine and cosine waves to produce a brand new wave. Consider the function y = cosx + sinx x 0  4  2 33 4 55 4 33 2 77 4  22 1 1 22 0 22 22 0 1 22 1 0 1 22 1 1 22 0 22 22 0 22-2-20011 1 cosx sinx cosx + sinx

4. The graph is like y = cosx, but it goes up to  2, and it has been moved to the right by  4 So we could write y = cosx + sinx as y =  2cos( x - )  4 The two waves have been combined to form a new wave.

5. Now here’s the maths bit… We can say, in general, kcos(x -  ) =k(cosxcos  + sinxsin  ) kcos(x -  ) =kcosxcos  + ksinxsin  kcos(x -  ) =(kcos  )cosx + (ksin  )sinx acosx + bsinx Given acosx + bsinx = kcos(x -  ) so, from original statement meaning: a = (kcos  ) and b = (ksin  ) = (kcos  )cosx + (ksin  )sinx

6. Now, a 2 =(kcos  ) 2 = k 2 cos 2  b 2 =(ksin  ) 2 = k 2 sin 2  a 2 + b 2 = k 2 (cos 2  + sin 2  ) a 2 + b 2 = k 2 k =  (a 2 + b 2 ) kcos  ksin  = tan  so Also,

7. So, simply acosx + bsinx = kcos(x -  ) k =  (a 2 + b 2 ) Where k and  can be obtained using tan  = kcos  ksin  a b =

8. EXAMPLE 1 Express 3cosxº + 4sinxº in the form kcos(x -  )º, where k>0 and 0<  <360. 3cosx + 4sinx = k(cosxcos  + sinxsin  ) 3cosx + 4sinx = kcosxcos  + ksinxsin  3cosx + 4sinx = (kcos  )cosx+(ksin  )sinx kcos  = 3 ksin  = 4 k =  (3 2 + 4 2 ) k =  25 k = 5

9. kcos  ksin  tan  = 3 4 = C A S T  = tan -1 3 4 () = 53.1 º  3cosxº+ 4sinxº =5cos(x-53.1)º

10. EXAMPLE 2 Express 5cosxº - 2sinxº in the form kcos(x -  )º, where k>0 and 0<  <360. 5cosx - 2sinx = k(cosxcos  + sinxsin  ) 5cosx - 2sinx = kcosxcos  + ksinxsin  5cosx - 2sinx = (kcos  )cosx+(ksin  )sinx kcos  = 5 ksin  = -2 k =  (5 2 + (-2) 2 ) k =  (25 + 4) k =  29

11. kcos  ksin  tan  = 5 -2 = C A S T tan -1 = 21.8 5 2 ( )  5cosxº- 2sinxº =  29cos(x-338.2)º  =360 - 21.8  = 338.2 º

12. EXAMPLE 3 Express 15cosxº - 8sinxº in the form kcos(x +  )º, where k>0 and 0<  <360. 15cosx - 8sinx = k(cosxcos  - sinxsin  ) 15cosx - 8sinx = kcosxcos  - ksinxsin  15cosx - 8sinx = (kcos  )cosx-(ksin  )sinx kcos  = 15 ksin  = 8 k =  (15 2 + (8) 2 ) k =  (225 + 64) k =  289 k = 17

13. kcos  ksin  tan  = 15 8 = C A S T  = tan -1 15 8 ()  15cosxº- 8sinxº =17cos(x+28.1)º  = 28.1 º

14. EXAMPLE 4 Express 7cosxº - 4sinxº in the form ksin(x +  )º, where k>0 and 0<  <360. 7cosx - 4sinx = k(sinxcos  + cosxsin  ) 7cosx - 4sinx = ksinxcos  + kcosxsin  7cosx - 4sinx = (kcos  )sinx+(ksin  )cosx kcos  = -4 ksin  = 7 k =  (-4) 2 + (7) 2 ) k =  (16 + 49) k =  65

15. kcos  ksin  tan  = -4 7 = C A S T tan -1 = 60.3 4 7 ()  7cosxº- 4sinxº =  65sin(x+119.7)º  = 180 – 60.3  = 119.7 º

16. Express 3cosxº - 5sinxº in the form ksin(x -  )º, where k>0 and 0<  <360. 3cosx - 5sinx = k(sinxcos  - cosxsin  ) 3cosx - 5sinx = ksinxcos  - kcosxsin  3cosx - 5sinx = (kcos  )sinx-(ksin  )cosx kcos  = -5 ksin  = -3 k =  (-5) 2 + (-3) 2 ) k =  (25 + 9) k =  34 EXAMPLE 5

17. kcos  ksin  tan  = -5 -3 = C A S T tan -1 = 31.0 5 3 ()  3cosxº- 5sinxº =  34sin(x–211.0)º  = 31.0 + 180  = 211.0 º

18. EXAMPLE 6 Express 6cos2xº - 8sin2xº in the form kcos(2x -  )º, where k>0 and 0<  <360. 6cos2x - 8sin2x = k(cos2xcos  + sin2xsin  ) 6cos2x - 8sin2x = kcos2xcos  + ksin2xsin  6cos2x - 8sin2x = (kcos  )cos2x+(ksin  )sin2x kcos  = 6 ksin  = -8 k =  (6) 2 + (-8) 2 ) k =  (36 + 64) k =  100 k = 10

19. kcos  ksin  tan  = 6 -8 = tan -1 = 51.3 6 8 ()  6cos2xº- 8sin2xº =10cos(2x-308.7)º  =360 - 51.3  = 308.7 º C A S T

20. Maximum and minimum values EXAMPLE 1 Find the maximum and minimum values of the function f(x) = 12sinx - 5cosx + 10, for 0º< x < 360º, and the corresponding values of x. Our first step is to express 12sinx - 5cosx as a single wave function 12sinx - 5cosx

21. 12sinx – 5cosx = k(cosxcos  + sinxsin  ) 12sinx – 5cosx = kcosxcos  + ksinxsin  12sinx – 5cosx = (kcos  )cosx+(ksin  )sinx kcos  = -5 ksin  = 12 k =  ((-5) 2 + 12 2 ) k =  (25 +144 ) k =  169 k = 13

22. cos  sin  tan  = -5 12 = tan -1 = 67.4º 5 12 ()  12sinx – 5cosx =13cos(x – 112.6) º  = 180 – 67.4  = 112.6 º C A S T f(x) = 12sinx – 5cosx + 10 f(x) = 13cos(x – 112.6) º + 10

23. QUESTIONWhat is the maximum value of f(x)? Think about the maximum value of cos(anything)? What is the maximum value of f(x)? Maximum value of cos (anything) is 1 Max. Value of f(x) = 13(1) + 10 = 23 when cos(x – 112.6) º = 1 (x – 112.6) º = 0, 360 x = 112.6 º

24. QUESTIONWhat is the minimum value of f(x)? Minimum value of cos (anything) is Min. Value of f(x) = 13(-1) + 10 = -3 when cos(x – 112.6) º = -1 (x – 112.6) º = 180, 540 x = 292.6 º f(x) = 13cos(x – 112.6) º + 10 Think about the minimum value of cos(anything)? What is the minimum value of f(x)?