Answer
a) $\dfrac{e^8-1}{2e^4}$; b) $\dfrac{15}{8}$
Work Step by Step
Apply definition of hyperbolic function for $\sinh x$
a) $\sinh 4=\dfrac{e^4-e^{-4}}{2}$
or, $=\dfrac{e^4-\frac{1}{e^4}}{2}$
or, $=\dfrac{\frac{e^8-1}{e^4}}{2}$
or, $=\dfrac{e^8-1}{2e^4}$
b) $\sinh ({\ln4})=\dfrac{e^{\ln 4}-e^{-({\ln 4})}}{2}$
or, $=\dfrac{4-\dfrac{1}{4}}{2}$
or, $=\dfrac{\dfrac{15}{4}}{2}$
or, $=\dfrac{15}{8}$
