# How to express $\cosh(4x)$ as a polynomial in $\cosh(x)$

My friend needs help with this question and I don't know it.

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If you use the definition of the hyperbolic cosine then, dropping the factor 2, you have:

$\exp(4x)+\exp(-4x) = (\exp(2x)-\exp(-2x))^2+2$

then,

$\exp(2x)-\exp(-2x) = (\exp(x)+\exp(-x))(\exp(x)-\exp(-x))$

if you do it you should get something like (if I'm not mistaken)

$\cosh(4x) = 8(\cosh(x))^2(\sinh(x))^2+1$

then, if you want to replace the $\sinh$ you can use the fact that $(\cosh(x))^2-(\sinh(x))^2=1$.

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$$\cosh(2x)=2\cosh^2(x)-1$$

$$\cosh(4x)=2\cosh^2(2x)-1=2(2\cosh^2(x)-1)^2-1=8\cosh^4(x)-8\cosh^2(x)+1$$

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For a mechanical way to translate the more familiar trigonometric identities to hyperbolic function identities, use $$\cosh x=\cos(ix),\qquad \sinh x=-i\sin(ix).$$

But perhaps an identity for $\cos(4w)$ does not qualify as familiar. However, it is easy to derive from standard facts: $$\cos(4w)=2\cos^2 (2w) -1=2(2\cos^2 w -1)^2-1.$$ Putting $w=ix$, we obtain in a mechanical way $$\cosh(4x)=2(2\cosh^2 x -1)^2-1.$$

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