# Power Reduction Formulas/Hyperbolic Cosine to 4th

## Theorem

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

where $\cos$ denotes hyperbolic cosine.

## Proof 1

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

$\blacksquare$

## Proof 2

 $\displaystyle \cosh^4 x$ $=$ $\displaystyle \frac 1 {2^4}\left(e^{x} + e^{-x}\right)^4$ Definition of Hyperbolic Cosine $\displaystyle$ $=$ $\displaystyle \frac 1 {16} \left({e^{4 x} + 4 e^{2 x} + 6 e^{0 x} + 4 e^{-2 x} + e^{-4 x} }\right)$ Binomial Theorem $\displaystyle$ $=$ $\displaystyle \frac 1 8 \left({\frac{e^{4 x} + e^{-4 x} } 2}\right) + \frac 4 8 \left({\frac{e^{2 x} + e^{-2 x} } 2 }\right) + \frac 6 {16}$ $\displaystyle$ $=$ $\displaystyle \frac {3 + 4 \cosh 2 x + \cosh 4 x} 8$ Definition of Hyperbolic Cosine

$\blacksquare$

## Proof 3

 $\displaystyle \cosh^4 x$ $=$ $\displaystyle \cos^4 i x$ Hyperbolic Cosine in terms of Cosine $\displaystyle$ $=$ $\displaystyle \frac {3 + 4 \cos \paren {2 i x} + \cos \paren {4 i x} } 8$ Fourth Power of Cosine $\displaystyle$ $=$ $\displaystyle \frac {3 + 4 \cosh 2 x + \cosh 4 x} 8$ Hyperbolic Cosine in terms of Cosine

$\blacksquare$