Use the definitions of $sin h x$ and $cos h x$ in terms of exponential functions to prove that $cos h 2 x = cos h^{2} x - 1$

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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;
LHS: $c o s h 2 x = \frac{e^{2 x} + e^{- 2 x}}{2}$
RHS: $2 c o s^{2} h x - 1 = 2 \times (\frac{e^{x} + e^{- x}}{2})^{2} - 1 =$ $2 \times (\frac{e^{2 x} + e^{- 2 x} + 2}{4}) - 1$
$\to$ $c o s h^{2} x - 1 =$ $\frac{e^{2 x} + e^{- 2 x}}{2}$
Thus;
$c o s h 2 x = c o s h^{2} - 1$

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