# Prove the following identity. ?

## Feb 21, 2018

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

So we rewrite the left-hand-side of the equation as follows:
$\sinh \left(a\right) + \sinh \left(b\right) = \frac{{e}^{a} - {e}^{-} a}{2} + \frac{{e}^{b} - {e}^{-} b}{2}$

This becomes:
$= \frac{{e}^{a} - {e}^{-} a + {e}^{b} - {e}^{-} b}{2}$
and rearranging gives:
$= \frac{{e}^{a} - {e}^{-} a + {e}^{b} - {e}^{-} b}{2}$ (Eq.A)

Now for the right-hand-side of the equation, we have:
$2 \sinh \left(\frac{a + b}{2}\right) \cosh \left(\frac{a - b}{2}\right) = 2 \left[\frac{{e}^{\frac{a + b}{2}} - {e}^{- \frac{a + b}{2}}}{2}\right] \left[\frac{{e}^{\frac{a - b}{2}} + {e}^{- \frac{a - b}{2}}}{2}\right]$
We can simplify this by cancelling out the $2$ and $\frac{1}{2}$ and keep only one $\frac{1}{2}$:
$= \frac{1}{2} \left[{e}^{\frac{a + b}{2}} - {e}^{- \frac{a + b}{2}}\right] \left[{e}^{\frac{a - b}{2}} + {e}^{- \frac{a - b}{2}}\right]$
we can also factor out ${e}^{\frac{1}{2}}$ as it appears everywhere:
$= {e}^{\frac{1}{2}} / 2 \left[{e}^{a + b} - {e}^{- a - b}\right] \left[{e}^{a - b} + {e}^{- a + b}\right]$
which becomes:
$= {e}^{\frac{1}{2}} / 2 \left[{e}^{a + b + a - b} + {e}^{a + b - a + b} - {e}^{- a - b + a - b} - {e}^{- a - b - a + b}\right]$
which simplifies to:
$= {e}^{\frac{1}{2}} / 2 \left[{e}^{2 a} + {e}^{2 b} - {e}^{- 2 b} - {e}^{- 2 a}\right]$
multiply back by ${e}^{\frac{1}{2}}$ and rearrange:
$= \frac{1}{2} \left[{e}^{a} - {e}^{- a} + {e}^{b} - {e}^{- b}\right]$
and you readily see that this is the same as our (Eq.A):
$= \frac{{e}^{a} - {e}^{- a}}{2} + \frac{{e}^{b} - {e}^{- b}}{2} = \sinh \left(a\right) + \sinh \left(b\right)$
So the left and right parts of the equation are the same and the identity is correct.

Q.E.D.