1. Prove equation

cosh2$\displaystyle theta$= 2sinh^2$\displaystyle theta$

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$\displaystyle theta$
I am trying to get the hang of this Latex

2. Originally Posted by Jaffa
cosh2$\displaystyle theta$= 2sinh^2$\displaystyle theta$

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$\displaystyle theta$
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

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

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