Show that $cos h x + sin h x = e^{x}$ and simplify $cos h x - sin h x$
By considering ${(cos h x + sin h x)}^{2} - {(cos h x - sin h x)}^{2}$ show that $2 sin h x cos h x = sin h 2 x$

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Solution

We know that;
$c o s h x = \frac{e^{x} + e^{- x}}{2}$

$s i n h x = \frac{e^{x} - e^{- x}}{2}$
Thus;
$c o s h x + s i n h x = \frac{e^{x} + e^{- x}}{2} + \frac{e^{x} - e^{- x}}{2} = e^{x}$

$c o s h x - s i n h x = \frac{e^{x} + e^{- x}}{2} - \frac{e^{x} - e^{- x}}{2} = e^{- x}$

Thus;
$(c o s h x + s i n h x)^{2} = e^{2 x}$

$(c o s h x - s i n h x)^{2} = e^{- 2 x}$

$\to$ $(c o s h x + s i n h x)^{2} -$ $(c o s h x - s i n h x)^{2} = e^{2 x} - e^{- 2 x}$

$\to 4 s i n h x c o s h x = e^{2 x} - e^{- 2 x}$

$\to 2 s i n h x c o s h x = \frac{e^{2 x} - e^{- 2 x}}{2}$

And
$s i n h 2 x = \frac{e^{2 x} - e^{- 2 x}}{2}$

$\to 2 s i n h x c o s h x$ $= s i n h 2 x$

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