# proving a hyperbolic trig identity

• Jan 23rd 2008, 02:44 PM
nmanik90
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
• Jan 23rd 2008, 03:15 PM
galactus
It's not that bad. Actually, it seems easier to prove than the standard trig identity.

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

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

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

$\displaystyle =sinh(x)cosh(y)+cosh(x)sinh(y)$
• Jan 23rd 2008, 03:26 PM
mr fantastic
Quote:

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

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

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

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

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

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

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 ;)
• Jan 23rd 2008, 03:28 PM
Plato
Here is another way to look at the question.
$\displaystyle \left( {e^x - e^{ - x} } \right)\left( {e^y + e^{ - y} } \right) = e^{x + y} + e^{x - y} - e^{-x + y} - e^{ - x - y}$
$\displaystyle \left( {e^x + e^{ - x} } \right)\left( {e^y - e^{ - y} } \right) = e^{x + y} - e^{x - y} + e^{ - x + y} - e^{ - x - y}$
Do you realize 2cosh(x)= $\displaystyle {e^x + e^{ - x} }$ and 2sinh(y)= $\displaystyle \left( {e^y - e^{ - y} } \right)$?