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Relation between Inverses of Trigonometric Functions and Their Reciprocal Functions

Use the defin...

Question

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

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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;
LHS: $c o s h 2 x = \frac{e^{2 x} + e^{- 2 x}}{2}$

RHS: $2 s i n h^{2} x + 1 = 2 \times (\frac{e^{x} - e^{- x}}{2})^{2} + 1 =$ $2 \times (\frac{e^{2 x} + e^{- 2 x} - 2}{4}) + 1$

$\to$ $s i n h^{2} x + 1 =$ $\frac{e^{2 x} + e^{- 2 x}}{2}$

Thus;
$c o s h 2 x = 1 + s i n^{2} h x$

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