Prove equation

**Jaffa** · March 18th 2009, 03:54 PM

cosh2 $theta$ = 2sinh^2 $theta$

I did this = 2.(e^theta -e^-theta)/2 x(e^theta -e^-theta)/2
=2 x 1/4(e^2theta + e^-2 theta)
=(e^2theta +e^-2theta)/2
= cosh2 $theta$
I am trying to get the hang of this Latex

**Danny** · March 18th 2009, 04:08 PM

Originally Posted by Jaffa

cosh2 $theta$ = 2sinh^2 $theta$

I did this = 2.(e^theta -e^-theta)/2 x(e^theta -e^-theta)/2 (1)
=2 x 1/4(e^2theta + e^-2 theta) (2)
=(e^2theta +e^-2theta)/2
= cosh2 $theta$
I am trying to get the hang of this Latex

Unfortunately, that's not the identity (and besides step (1) to (2) is wrong - there's a cross term missing). The identity is

$\cosh 2 x = \cosh^2 x + \sinh^2 x$

Proof:
$\cosh^2 x + \sinh^2 x = \left( \frac{e^x + e^{-x} }{2} \right) ^2 + \left( \frac{e^x - e^{-x} }{2} \right) ^2 = \frac{e^{2x} + 2 + e^{-2x}}{4} + \frac{e^{2x} - 2 + e^{-2x}}{4}$
$= \frac{e^{2x} + e^{-2x}}{2} = \cosh 2x$

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