Prove that cosh²x – sinh²x = 1

by RoRi

November 17, 2015

Prove that

$\cosh^2 x - \sinh^2 x = 1.$

Proof. We use the definitions of $\cosh x$ and $\sinh x$ in terms of the exponential function,

$\begin{align*} \cosh^2 x - \sinh^2 x &= \left( \frac{e^x + e^{-x}}{2} \right)^2 - \left( \frac{e^x - e^{-x}}{2} \right)^2 \\ &= \frac{e^{2x} + 2 + e^{-2x} - e^{2x} + 2 - e^{-2x}}{4} \\ &= \frac{4}{4} \\ &= 1. \qquad \blacksquare \end{align*}$

3 comments

February 24, 2021 at 10:33 pm

Charles says:

sech(x)*e^x=2+2e^(2x)

Simple Identity, but I can’t seem to prove it

Reply
November 2, 2017 at 6:05 am

nadun says:

thanks verymuch

Reply
December 18, 2016 at 4:11 am

Anonymous says:

Thank you sir

Reply

Stumbling Robot

Prove that cosh²x – sinh²x = 1

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