$\cosh^4x-\sinh^4x=\cosh2x$

Question

I need to show that $\cosh^4(x)-\sinh^4(x) = \cosh(2x)$

First I found myself going in circles..

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

Now I'm trying to get somewhere using the identity $$\cosh ^2(x)-\sinh ^2(x)=1$$

if $\cosh ^2(x)-\sinh ^2(x)=1$ then $$\cosh ^2(x)-\sinh ^2(x)=\left(\frac{1}{2} \left(e^{-x}+e^x\right)\right)^2-\left(\frac{1}{2} \left(e^x-e^{-x}\right)\right)^2$$

yet the same doesn't apply when I take them to the $4^\text{th}$ power.

Please could someone point me in the right direction as I'm getting very lost here.

Answer 1

Hint: Factor $$ (\cosh x )^4 -(\sinh x)^4 = ((\cosh x )^2 -(\sinh x)^2 )(\cosh x )^2 +(\sinh x)^2 ) \\= (\cosh x )^2 +(\sinh x)^2 $$

and then use identites for $\cosh(2x)$

Answer 2

HINT: You know that

$$\begin{align*} \cosh^4x-\sinh^4x&=\left(\cosh^2x-\sinh^2x\right)\left(\cosh^2x+\sinh^2x\right)\\ &=\cosh^2x+\sinh^2x\\ &=\left(\frac{e^x+e^{-x}}2\right)^2+\left(\frac{e^x-e^{-x}}2\right)^2\;; \end{align*}$$

can you finish it from there?

Answer 3

the left hand side is given by $$2e^{2x}+2e^{-2x}$$ and the right hand side is $$2e^{2x}+2e^{-2x}$$ thus our equation is true for all real $$x$$