Hyperbolic Cosine minus Hyperbolic Cosine/Proof 2
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
\(\ds 2 \map \sinh {\dfrac {x + y} 2} \map \sinh {\dfrac {x - y} 2}\) | \(=\) | \(\ds \dfrac 1 2 \paren {e^{\frac {x + y} 2} - e^{-\frac {x + y} 2} } \paren {e^{\frac {x - y} 2} - e^{-\frac {x - y} 2} }\) | Definition of Hyperbolic Sine | |||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac 1 2 \paren {e^x - e^y - e^{-y} + e^{-x} }\) | simplifying | |||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {e^x + e^{-x} } 2 - \dfrac {e^y + e^{-y} } 2\) | rearranging | |||||||||||
\(\ds \) | \(=\) | \(\ds \cosh x - \cosh y\) | Definition of Hyperbolic Cosine |
$\blacksquare$