Single Variable Calculus, Chapter 7, 7.7, Section 7.7, Problem 10

Show that $\cos h x - \sin h x = e^{-x}$

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

\end{aligned}
\end{equation}
$