Single Variable Calculus, Chapter 7, 7.7, Section 7.7, Problem 12

Show that $\cos h(x + y) = \cos h x \cos h y + \sin h x \sin h y$

Solving for the left-hand side of the equation

Using Hyperbolic Function


$
\begin{equation}
\begin{aligned}

\cos h x =& \frac{e^x + e^{-x}}{2}
\\
\\
\cos h (x + y) =& \frac{e^{(x + y)} + e^{- (x + y)} }{2}
\\
\\
\cos h (x + y) =& \frac{e^{(x + y)} + e^{(-x-y)} }{2}
\\
\\
\cos h (x + y) =& \frac{e^x e^y + e^{-x} e^{-y}}{2}

\end{aligned}
\end{equation}
$


Using the Identities


$
\begin{equation}
\begin{aligned}

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


\end{aligned}
\end{equation}
$