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Maximum of

  1. Jun 23, 2006 #1
    The question is find the maximum value of the following function

    f(x) = 3cos(4*pi*x-1.3) + 5cos(2*pi*x+0.5).
     
  2. jcsd
  3. Jun 24, 2006 #2

    HallsofIvy

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    "The question"? Where, in your homework?

    Looks pretty straight forward to me: differentiate and set the derivative equal to 0. Use the sin(a+b) formula to isolate [itex]sin(3\pi x)[/itex] and [itex]cos(3\pi x)[/itex].
     
  4. Jun 25, 2006 #3
    Calculus isn't always helpful!

    Hey HallsofIvy,

    Your suggestion was quite correct. However I dare say differentiating and equating to zero does not always help.
    Here f (x) = 3cos (4*pi*x-1.3) + 5cos (2*pi*x+0.5)
    Or, f’ (x) = - [12*pi*sin (4*pi*x-1.3) + 10*pi*sin (2*pi*x+0.5)]
    Equating this to zero,
    12*pi*sin (4*pi*x-1.3) + 10*pi*sin (2*pi*x+0.5)] = 0
    Now if I use the sin (a+b) formula the equation gets rather complicated.

    I shall be thankful to you if you could please show the full solution.
    However the answer is 5.7811.
     
  5. Jun 25, 2006 #4

    HallsofIvy

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    Yes, it gets complicated. Anything wrong with that? You'll also, by the way, need to use the double angle formulas to reduce [itex]4\pi x[/itex] to [itex]2\pi x[/itex]. After you done all that, you will have an equation of the form [itex]A sin(2\pi x)+ B cos(2\pi x)= 0[/itex] with rather complicated numbers for A and B. But they are only numbers! Write [itex]tan(2\pi x)= -B/A[/itex] and solve.
     
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