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Alternative to solve this?

  1. Nov 1, 2012 #1
    I need to prove that if ##\sin(x+iy)=cos(\alpha)+i\sin(\alpha)## , then ##\cosh(2y)-\cos(2x)=2##

    expanding sin(x+iy) and equating real and imaginary parts, we get:
    ##\cos(\alpha)=\sin(x)\cosh(y)## -- (1)
    and ## \sin(\alpha)=\cos(x)\sinh(y) ## -- (2)

    if we use ##\cos^2(\alpha)+\sin^2(\alpha)=1 ## and put the values from (1) and (2) , we can get the required expression. My question is that, is there an alternative to do this? Can I prove it by directly using the values of cosh2y and cos2x [from (1) and (2)]? I tried but it seems impossible. Why is it so?
     
  2. jcsd
  3. Nov 1, 2012 #2

    lurflurf

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    use
    cosh(a)=cos(i a)
    then
    cos(2a)-cos(2b)=-2sin(a-b)sin(a+b)
    and so forth
     
  4. Nov 1, 2012 #3
    ok..what is a and b?
     
  5. Nov 1, 2012 #4

    lurflurf

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    Whatever you want (those are identities), in this case perhaps
    cosh(2y)=cos(2i y)
    cos(2iy)-cos(2x)=-2sin(iy-x)sin(iy+x)
     
  6. Nov 1, 2012 #5
    Ok that is brilliant. thanks!
     
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