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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- Mar 21st 2010, 09:19 PMferoniusDerive 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! - Mar 21st 2010, 10:51 PMmatheagle
Just substitute the definitions and multiply.

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

likewise....

$\displaystyle \sinh x \cosh y= \left({e^x-e^{-x}\over 2}\right)\left({e^y+e^{-y}\over 2}\right)$ - Mar 23rd 2010, 07:01 PMferonius
thanks, didnt realise it was as simple as that, muchly appreciated!