# Math Help - proving a hyperbolic trig identity

1. ## proving a hyperbolic trig identity

how would you prove that:

sinh (x+y) = sinhx coshy + coshx sinhy ??

this is a little bit tricky (for me!) since it involves hyperbolic functions

2. It's not that bad. Actually, it seems easier to prove than the standard trig identity.

Use $cosh(x)+sinh(x)=e^{x}, \;\ cosh(x)-sinh(x)=e^{-x}$

$sinh(x+y)=\frac{e^{x+y}-e^{-(x+y)}}{2}=\frac{e^{x}e^{y}-e^{-x}e^{-y}}{2}$

$=\frac{1}{2}[(cosh(x)+sinh(x))(cosh(y)+sinh(y))-(cosh(x)-sinh(x))$ $(cosh(y)-sinh(y))]$

$=sinh(x)cosh(y)+cosh(x)sinh(y)$

3. Originally Posted by nmanik90
how would you prove that:

sinh (x+y) = sinhx coshy + coshx sinhy ??

this is a little bit tricky (for me!) since it involves hyperbolic functions
1. $\displaystyle \sinh (x + y) = \frac{1}{2} \left( e^{x + y} - e^{-(x + y)}\right) = \frac{1}{2} \left( e^{x + y} - e^{-x - y} \right)$

2. $\displaystyle e^{x + y} - e^{-x - y} = \frac{1}{2} \left( 2e^{x + y} - 2e^{-x - y} \right)$

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

$\displaystyle = \frac{1}{2}\left( e^{x + y} + e^{x + y} + e^{x - y} - e^{x - y} + e^{-x + y} - e^{-x + y} + e^{-x - y)} + e^{-x - y}\right)$

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

$\displaystyle \frac{1}{2} \left( e^x - e^{-x}\right) \left(e^y + e^{-y} \right) + \frac{1}{2} \left( e^x + e^{-x}\right) \left(e^y - e^{-y} \right)$

So, putting these two results together ........

In answer to your question "How did you know to do that?", it helps to know what the answer has to be

4. Here is another way to look at the question.
$\left( {e^x - e^{ - x} } \right)\left( {e^y + e^{ - y} } \right) = e^{x + y} + e^{x - y} - e^{-x + y} - e^{ - x - y}$
$\left( {e^x + e^{ - x} } \right)\left( {e^y - e^{ - y} } \right) = e^{x + y} - e^{x - y} + e^{ - x + y} - e^{ - x - y}$
Do you realize 2cosh(x)= ${e^x + e^{ - x} }$ and 2sinh(y)= $\left( {e^y - e^{ - y} } \right)$?