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Thread: Derive the identity

  1. March 21st 2010, 09:19 PM #1
    feronius
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    Derive the identity

    hey just having a little trouble with this problem,

    Derive the identity:
    sinh x cosh y = (1/2)[sinh(x+y)+sinh(x-y)]

    thanks!
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  2. March 21st 2010, 10:51 PM #2
    matheagle
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    Just substitute the definitions and multiply.

    (1/2)[\sinh(x+y)+\sinh(x-y)]=(1/4)\left[ e^{x+y}-e^{-x-y}+e^{x-y}-e^{-x+y}\right]

    likewise....

    \sinh x \cosh y= \left({e^x-e^{-x}\over 2}\right)\left({e^y+e^{-y}\over 2}\right)
    Last edited by matheagle; March 23rd 2010 at 08:10 PM.
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  3. March 23rd 2010, 07:01 PM #3
    feronius
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    thanks, didnt realise it was as simple as that, muchly appreciated!
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