1. ## 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

2. Hello,
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 $\displaystyle (x^a)^b=x^{ab}$

3. 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!

4. *facepalms*

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

Thanks guys