Prosthaphaeresis Formulas/Hyperbolic Cosine minus Hyperbolic Cosine/Proof 2

From ProofWiki
Jump to navigation Jump to search

Theorem

$\cosh x - \cosh y = 2 \, \map \sinh {\dfrac {x + y} 2} \, \map \sinh {\dfrac {x - y} 2}$


Proof

\(\displaystyle 2 \, \map \sinh {\dfrac {x + y} 2} \, \map \sinh {\dfrac {x - y} 2}\) \(=\) \(\displaystyle \dfrac 1 2 \paren {e^{\frac {x + y} 2} - e^{-\frac {x + y} 2} } \paren {e^{\frac {x - y} 2} - e^{-\frac {x - y} 2} }\) Definition of Hyperbolic Sine
\(\displaystyle \) \(=\) \(\displaystyle \dfrac 1 2 \paren {e^x - e^y - e^{-y} + e^{-x} }\) simplifying
\(\displaystyle \) \(=\) \(\displaystyle \dfrac {e^x + e^{-x} } 2 - \dfrac {e^y + e^{-y} } 2\) rearranging
\(\displaystyle \) \(=\) \(\displaystyle \cosh x - \cosh y\) Definition of Hyperbolic Cosine

$\blacksquare$