Chapter 3.2, Problem 60E

Solution

To prove: That the hyperbolic cosine function is an even function.

The hyperbolic cosine function is an even function.

Given:

  $\cosh \left(x\right)=\frac{e^x+e^{-x}}{2}$ .

Concept used:

A function $f\left(x\right)$ is said to be an even function if $f\left(-x\right)=f\left(x\right)$ .

Calculation:

Consider,

  $\begin{array}{c}\cosh \left(-x\right)=\frac{e^{-x}+e^{-\left(-x\right)}}{2}\\ =\frac{e^{-x}+e^x}{2}\\ =\frac{e^x+e^{-x}}{2}\\ =\cosh \left(x\right)\end{array}$

Therefore, $\cosh \left(x\right)$ is an even function.