Answer
a) 0 b) 1
Work Step by Step
a) Since, $\sinh(x)$ = $\dfrac{e^x-e^-x}{2}$
Thus, $\sinh(0)=\dfrac{(e^0)-(e^-0)}{2}=\dfrac{1-1}{2}=0$
b) Since, $\cosh(x)$ = $\dfrac{e^x+e^{-x}}{2}$
Thus, $\cosh(0)$= $\dfrac{(e^0)+(e^-0)}{2}$=$\dfrac{1+1}{2}=1$
