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

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.

• Please consider using \cosh and \sinh to get $\cosh$ and $\sinh$ instead of cosh and sinh which give $cosh$ and $sinh$. Commented Aug 7, 2015 at 19:17
• To "Fly by Night"'s comment I would add that the backslash not only prevents italicization but also results in proper spacing in things like $a\cosh b$. ${}\qquad{}$ Commented Aug 7, 2015 at 19:25

## 3 Answers

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 $\cosh(2x)$

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?

• doesn't $(\frac{e^x + e^-x}{2})^2 + \frac{e^x - e^-x}{2})^2 = 2cosh(2x)$? @Brian M. Scott Commented Aug 8, 2015 at 22:13
• @Mathsguy9020: No, it's $$\frac{1}{4}\left(e^{2x}+2+e^{-2x}+e^{2x}-2+e^{-2x}\right)=\frac{e^{2x}+e^{-2x}}2=\cosh 2x\;.$$ Commented Aug 8, 2015 at 23:50

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 $$x$$