Proving Hyperbolic Equation

Question

I want to prove the following equality (equation): \begin{equation} (c\cosh(x) + c\sinh)^2 e^{-2x} = (c\cosh(x))^2 - (c\sinh(x))^2 \end{equation} where c is just a constant So i need to grab one side of the equation(one expression) and try to get to the other (or use an other method, like take the entire equation and try to reach an equality between both sides of the equation). Anyway i took the first part of the equation and i know that \begin{equation} e^{-x} = \cosh(x) - \sinh(x) \end{equation} so i've reached at this point (by also taking out c from the first parenthesis): \begin{equation} c^2(\cosh(x)^2 - \sinh(x)^2)^2 \end{equation} and i'm stuck. I don't know how to prove it. If anyone could help please.. Thanks a lot in advance.

Answer 1

Going from left to right $$ c^2\left(\cosh (x) +\sinh(x)\right)^2\mathrm{e}^{-2x} = c^2\left(\cosh (x) +\sinh(x)\right)^2\left(\cosh (x) -\sinh(x)\right)^2 = c^2\left(\cosh^2 x-\sinh ^2 x\right)^2 $$ now we have an identity (this should be ingrained) $$ \cosh^2 x-\sinh ^2 x = 1 $$ thus the lhs side is $$ c^2\left(\cosh (x) +\sinh(x)\right)^2\mathrm{e}^{-2x} = c^2 $$ it is trivial to prove that the other side is $c^2$.

Answer 2

There is a $c^2$ in both sides of your equation. There is no need for it. Instead, just see $e^x = \cosh x + \sinh x$ and $e^{-x} = \cosh h - \sinh x$ thus $$ 1=e^0=e^{x-x}=e^xe^{-x} = (\cosh x + \sinh x)(\cosh h - \sinh x) = \cosh^2 x - \sinh^2 x.$$ done.