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Complex - Trig Identity

  1. Sep 16, 2006 #1
    I am asked to prove the following: (Note: z = x + iy)

    |cos(z)|2 = cos2x + sinh2y
    ---------------

    So I started the following way:

    |cos(z)|2 = |cos(x+iy)|2
    = |cos(x)cosh(y) - i(sin(x)sinh(y))|2
    = cos2(x)cosh2(y) + sin2(x)sinh2(y) [after having square root squared removed]

    once I got here I was stuck. I am just not seeing how we can get this to equal cos2x + sinh2y

    Is there some silly trig identity I don't know? Or did I make a mistake? Any ideas?

    Thanks!
     
    Last edited: Sep 16, 2006
  2. jcsd
  3. Sep 16, 2006 #2

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    I think you're right and the given answer is wrong. Try plugging in a few values of z to check.
     
  4. Sep 16, 2006 #3
    I thought about it maybe being wrong too, never thought of pluging in values though, duh! :smile:

    Take [itex]z = \pi / 2[/itex] and clearly we get something different on both sides. Guess I will talk to my professor about it since he wrote the homework himself. Thanks!
     
    Last edited: Sep 16, 2006
  5. Sep 18, 2006 #4
    Actually, my z = pi/2 is not a counterexample (since x = pi/2 and y = 0, we get 0 on both sides).

    For the actual proof, I just need to continue where I left off, but change sin2 to 1-cos2, which supposedly gets what we want (I have yet to work it out).
     
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