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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.

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  • $\begingroup$ Use this identity that $a^2-b^2=(a-b)(a+b)$ in the right hand of your first equation ... $\endgroup$ – Math-fun Jul 29 '15 at 13:53
  • $\begingroup$ Well i did that, but it still isn't equal with the left side. $\endgroup$ – Nikos Jul 29 '15 at 13:58
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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 $c^2$.

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  • $\begingroup$ I knew this identity but obviously i was too stupid to think about it, i was stuck. Thank you all very much. $\endgroup$ – Nikos Jul 29 '15 at 14:11
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    $\begingroup$ Not too stupid! This question made me go down a path that I didnt want to go down..before realizing that was the crux of the answer. $\endgroup$ – Chinny84 Jul 29 '15 at 14:14
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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.

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    $\begingroup$ btw, $\cosh x = \frac{1}{2}(e^x+e^{-x})$ and $\sinh x = \frac{1}{2}(e^x-e^{-x})$ can be seen as the even+odd decomposition of the function $e^x$ as it is manifestly clear that $e^x = \frac{1}{2}(e^x+e^{-x})+\frac{1}{2}(e^x-e^{-x})$. $\endgroup$ – James S. Cook Jul 29 '15 at 14:12

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