Power Reduction Formulas/Hyperbolic Sine to 4th

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Theorem

$\sinh^4 x = \dfrac {3 - 4 \cosh 2 x + \cosh 4 x} 8$

where $\sinh$ and $\cosh$ denote hyperbolic sine and hyperbolic cosine respectively.


Proof

\(\ds \sinh^4 x\) \(=\) \(\ds \left({\sinh^2 x}\right)^2\)
\(\ds \) \(=\) \(\ds \left({\frac {\cosh 2 x - 1} 2}\right)^2\) Square of Hyperbolic Sine
\(\ds \) \(=\) \(\ds \frac {\cosh^2 2 x - 2 \cosh 2 x + 1} 4\) multiplying out
\(\ds \) \(=\) \(\ds \frac {\frac {\cosh 4 x + 1} 2 - 2 \cosh 2 x + 1} 4\) Square of Hyperbolic Cosine
\(\ds \) \(=\) \(\ds \frac {\cosh 4 x + 1 - 4 \cosh 2 x + 2} 8\) multiplying top and bottom by $2$
\(\ds \) \(=\) \(\ds \frac {3 - 4 \cosh 2 x + \cosh 4 x} 8\) rearrangement

$\blacksquare$


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