Difference of Squares of Hyperbolic Cosine and Sine

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Theorem

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

where $\cosh$ and $\sinh$ are hyperbolic cosine and hyperbolic sine.


Proof

\(\ds \cosh^2 x - \sinh^2 x\) \(=\) \(\ds \paren {\frac {e^x + e^{-x} } 2}^2 - \paren {\frac {e^x - e^{-x} } 2}^2\) Definition of Hyperbolic Cosine and Definition of Hyperbolic Sine
\(\ds \) \(=\) \(\ds \paren {\frac {\paren {e^x}^2 + 2 \paren {e^x} \paren {e^{-x} } + \paren {e^{-x} }^2} 4} - \paren {\frac {\paren {e^x}^2 - 2 \paren {e^x} \paren {e^{-x} } + \paren {e^{-x} }^2} 4}\) Square of Sum
\(\ds \) \(=\) \(\ds \paren {\frac {e^{2 x} + 2 + e^{-2 x} } 4} - \paren {\frac {e^{2 x} - 2 + e^{-2 x} } 4}\) Exponential of Sum
\(\ds \) \(=\) \(\ds \frac {e^{2 x} - e^{2 x} + e^{-2 x} - e^{-2 x} + 2 + 2} 4\)
\(\ds \) \(=\) \(\ds 1\)

$\blacksquare$


Also presented as

This result can also be presented as:

$\cosh^2 x = 1 + \sinh^2 x$

or:

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


Also see


Sources