May 28, 2021 - 11:34

An infinite series problem.

By Vamshi Jandhyala in mathematics

May 28, 2021

Problem

Solution

We have

$$ \begin{aligned} \cosh{2x} &= 2\cosh^2{x} - 1 \\
\cosh{2x} &= 1 + 2\sinh^2{x} \\
\cos{2x} &= 2\cos^2{x} - 1 \\
\cos{2x} &= 1 - 2\sin^2{x}\\
\end{aligned} $$

Therefore,

$$ \begin{aligned} \sin^2{x}\cosh^2{x} + \cos^2{x}\sinh^2{x} &= \frac{(1 - \cos{2x})(1 + \cosh{2x})}{2} + \frac{(1 + \cos{2x})(\cosh{2x}-1)}{2} \\
&= \frac{\cosh{2x} - \cos{2x}}{2} \end{aligned} $$