# Hyperbolic Identity problem

• Mar 22nd 2009, 03:18 AM
Stonehambey
Hyperbolic Identity problem
Wasn't too sure where to put this one, so feel free to yell at me if it's in the wrong forum. :)

I'm having some difficulty starting this problem,

Show that

$\displaystyle (\cosh x + \sinh x)^k + (\cosh x - \sinh x)^k = 2\cosh kx$

I thought about proof by induction, or expanding the left hand side using the binomial theorem. But both times I got stuck. Any help would be much appreciated. Thanks

Stonehambey
• Mar 22nd 2009, 04:01 AM
Moo
Hello,
Quote:

Originally Posted by Stonehambey
Wasn't too sure where to put this one, so feel free to yell at me if it's in the wrong forum. :)

I'm having some difficulty starting this problem,

Show that

$\displaystyle (\cosh x + \sinh x)^k + (\cosh x - \sinh x)^k = 2\cosh kx$

I thought about proof by induction, or expanding the left hand side using the binomial theorem. But both times I got stuck. Any help would be much appreciated. Thanks

Stonehambey

Here are some big hints :
$\displaystyle \cosh(x)=\frac{e^x+e^{-x}}{2}$
$\displaystyle \sinh(x)=\frac{e^x-e^{-x}}{2}$
Thus $\displaystyle \cosh(x)+\sinh(x)=\dots$ and $\displaystyle \cosh(x)-\sinh(x)=\dots$

And remember that $\displaystyle (x^a)^b=x^{ab}$
• Mar 22nd 2009, 04:05 AM
Hush_Hush
You only have to put in equation this definitions:

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

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

$\displaystyle (cosh(x)+sinh(x))^k+(cosh(x)-sinh(x))^k=e^{kx}+e^{-kx}=2cosh(kx)$

That's all,

Have a nice day!
• Mar 22nd 2009, 04:44 AM
Stonehambey
*facepalms*

Talk about not being able to see the wood for the trees!

Thanks guys :)