# Chapter 3 - Section 3.11 - Hyperbolic Functions - 3.11 Exercises: 1

(a) 0 (b) 1

#### Work Step by Step

(a): recall $sinh(x)$ = $\frac{(e^x)-(e^-x)}{2}$; therefore, $sinh(0)$= $\frac{(e^0)-(e^-0)}{2}$=$\frac{(1)-(1)}{2}$=$\frac{0}{2}$=$0$ (b): recall $cosh(x)$ = $\frac{(e^x)+(e^-x)}{2}$; therefore, $cosh(0)$= $\frac{(e^0)+(e^-0)}{2}$=$\frac{(1)+(1)}{2}$=$\frac{2}{2}$=$1$

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.