Power Reduction Formulas/Hyperbolic Cosine to 4th

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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

$\blacksquare$


Also see


Sources