How can I demostrate the $\sinh^{4} x$ identity?

Question

How can I prove $\sinh^{4}x=\frac{1}{8}\left(\cosh 4x-4\cosh 2x+3\right)$?

Answer 1

$\sinh^4 x = (\frac{1}{2}e^x - \frac{1}{2}e^{-x})^4= \frac{1}{16}(e^x - e^{-x})^4$ which by Newton's formula equals

$\frac{1}{16}\left(e^{4x} - \binom{4}{1}e^{3x}e^{-x} + \binom{4}{2}e^{2x}e^{-2x} - \binom{4}{3}e^x e^{-3x} + e^{-4x}\right)$

which becomes

$\frac{1}{16}\left(e^{4x} - 4e^{2x} + 6 - 4e^{-2x} + e^{-4x}\right)$

And use now that $e^{4x}+e^{-4x}=2\cosh (4x)$ by definition and $e^{2x}+e^{-2x} = 2\cosh 2x$ as well.

So we get

$\frac{1}{16}\left(2\cosh 4x - 8\cosh 2x + 6\right)$ and bringing the $2$'s from inside the braces to outside, we get your target result

$\frac{1}{8}\left(\cosh 4x - 4\cosh 2x + 3 \right)$

Answer 2

Hint

$$2\sinh(x)=e^x-e^{-x}$$

Use binomial expansion

Then replace $$e^y+e^{-y}$$ with $$2\cosh(y)$$ where $y=2x,4x$