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!
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....
$\displaystyle \sinh x \cosh y= \left({e^x-e^{-x}\over 2}\right)\left({e^y+e^{-y}\over 2}\right)$