# Prove the following identity. ?

Feb 21, 2018

The identity is:
$\cosh \left(a\right) + \cosh \left(b\right) = 2 \cosh \left(\frac{a + b}{2}\right) \cosh \left(\frac{a - b}{2}\right)$
See explanation for its derivation.

#### Explanation:

You must know the definition of $\sinh$ and $\cosh$ to find the identity.
By definition:
$\sinh x = \frac{{e}^{x} - {e}^{-} x}{2}$ and
$\cosh x = \frac{{e}^{x} + {e}^{-} x}{2}$

Hint: see how is the identity of $\sinh a + \sinh b$.

We start off by rewriting:
$\cosh a + \cosh b = \frac{{e}^{a} + {e}^{-} a}{2} + \frac{{e}^{b} + {e}^{-} b}{2}$

factoring out $\frac{1}{2}$ gives:
$= \frac{1}{2} \left[{e}^{a} + {e}^{-} a + {e}^{b} + {e}^{-} b\right]$

factoring out ${e}^{\frac{1}{2}}$ gives:
$= {e}^{\frac{1}{2}} / 2 \left[{e}^{2 a} + {e}^{- 2 a} + {e}^{2 b} + {e}^{- 2 b}\right]$

Trick! Add "zeroes" in the form of $a - a$ or $b - b$:
$= {e}^{\frac{1}{2}} / 2 \left[{e}^{a + a + b - b} + {e}^{- a - a + b - b} + {e}^{b + b + a - a} + {e}^{- b - b + a - a}\right]$

Separate $a$ from $a$ and $b$ from $b$ (and rearrange) to get :
$= {e}^{\frac{1}{2}} / 2 \left[{e}^{a + b} {e}^{a - b} + {e}^{-} \left(a + b\right) {e}^{-} \left(a - b\right) + {e}^{a + b} {e}^{-} \left(a - b\right) + {e}^{-} \left(a + b\right) {e}^{a - b}\right]$

This becomes:
$= {e}^{\frac{1}{2}} / 2 \left[\left({e}^{a + b} + {e}^{-} \left(a + b\right)\right) \left({e}^{a - b} + {e}^{-} \left(a - b\right)\right)\right]$

Re-introduce ${e}^{\frac{1}{2}}$
$= \frac{1}{2} \left[\left({e}^{\frac{a + b}{2}} + {e}^{- \frac{a + b}{2}}\right) \left({e}^{\frac{a - b}{2}} + {e}^{- \frac{a - b}{2}}\right)\right]$

Trick! Multiply by $1 = \frac{2}{2}$:
$= 2 \frac{\left({e}^{\frac{a + b}{2}} + {e}^{- \frac{a + b}{2}}\right)}{2} \frac{\left({e}^{\frac{a - b}{2}} + {e}^{- \frac{a - b}{2}}\right)}{2}$
which is really just
$= 2 \cosh \left(\frac{a + b}{2}\right) \cosh \left(\frac{a - b}{2}\right)$

The identity is:
$\cosh \left(a\right) + \cosh \left(b\right) = 2 \cosh \left(\frac{a + b}{2}\right) \cosh \left(\frac{a - b}{2}\right)$

Quite tricky!