Single Variable Calculus, Chapter 7, 7.7, Section 7.7, Problem 16

Show that $\displaystyle \cos h 2x = \cos h^2 x + \sin h^2 x$

Solving for the left-hand side of the equation

Using Hyperbolic Function


$
\begin{equation}
\begin{aligned}

\cos hx =& \frac{e^x + e^{-x}}{2}
\\
\\
\cos h 2 x =& \frac{e^{2x} + e^{-2x}}{2}
\\
\\
\cos h 2 x =& \frac{e^x e^x + e^{-x} e^{-x}}{2}

\end{aligned}
\end{equation}
$


Using Hyperbolic Identities


$
\begin{equation}
\begin{aligned}

& \cos hx + \sin hx = e^x \text{ and } \cos hx - \sin hx = e^{-x}
\\
\\
& \cos h 2 x = \frac{(\cos hx + \sin hx)(\cos hx + \sin hx) + (\cos hx - \sin hx)(\cos hx - \sin hx)}{2}
\\
\\
& \cos h 2 x = \frac{(\cos hx + \sin hx)^2 + (\cos hx - \sin hx)^2}{2}
\\
\\
& \cos h 2 x = \frac{\cos h^2 x + \cancel{2 \cos hx \sin hx} + \sin h^2 x + \cos h^2 x - \cancel{2 \cos hx \sin hx} + \sin h^2 x}{2}
\\
\\
& \cos h 2 x = \frac{2 \cos h^2 x + 2 \sin h^2 x}{2}
\\
\\
& \cos h 2 x = \frac{\cancel{2} (\cos h^2 x + \sin h^2 x)}{\cancel{2}}
\\
\\
& \cos h 2 x = \cos h^2 x + \sin h^2 x



\end{aligned}
\end{equation}
$

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