# How do you prove Cosh(A+B) = CoshA*CoshB+SinhB*SinhA?

May 3, 2016

See below

#### Explanation:

Use the definition of Hyperbolic Functions for cosh x
$\cosh x = \frac{{e}^{x} + {e}^{-} x}{2}$ and $\sinh x = \frac{{e}^{x} - {e}^{-} x}{2}$

$\cosh \left(A + B\right) = \cosh A \cosh B + \sinh A \sinh B$

Right Side: $= \left[\frac{{e}^{A} + {e}^{-} A}{2} \times \frac{{e}^{B} + {e}^{-} B}{2}\right] + \left[\frac{{e}^{A} - {e}^{-} A}{2} \times \frac{{e}^{B} - {e}^{-} B}{2}\right]$

$= \frac{{e}^{A + B} + {e}^{A - B} + {e}^{B - A} + {e}^{- \left(A + B\right)}}{4} + \frac{{e}^{A + B} - {e}^{A - B} - {e}^{B - A} + {e}^{- \left(A + B\right)}}{4}$

$= \frac{{e}^{A + B} + \cancel{{e}^{A - B}} + \cancel{{e}^{B - A}} + {e}^{- \left(A + B\right)} + {e}^{A + B} - \cancel{{e}^{A - B}} - \cancel{{e}^{B - A}} + {e}^{- \left(A + B\right)}}{4}$

$= \frac{2 {e}^{A + B} + 2 {e}^{- \left(A + B\right)}}{4}$

$= \frac{2 \left({e}^{A + B} + {e}^{- \left(A + B\right)}\right)}{4}$

$= \frac{{e}^{A + B} + {e}^{- \left(A + B\right)}}{2}$

$= \cosh \left(A + B\right)$

$=$right side