Power Reduction Formulas/Hyperbolic Cosine to 4th/Proof 1

Theorem

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

$\ds \cosh 4 x$

$=$

$\ds \paren {\cosh^2 x}^2$

$\ds $

$=$

$\ds \paren {\frac {\cosh 2 x + 1} 2}^2$

$\ds $

$=$

$\ds \frac {\cosh^2 2 x + 2 \cosh 2 x + 1} 4$

multiplying out

$\ds $

$=$

$\ds \frac {\frac {\cosh 4 x + 1} 2 + 2 \cosh 2 x + 1} 4$

$\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

$\blacksquare$