Answer
$f(\theta)=-cos(\theta)+6$
Work Step by Step
$f’’(\theta)=cos(\theta),$ $f’(\frac{\pi}{2})=1,$ $f(\frac{\pi}{2})=6$
$f’(\theta)=\int cos(\theta)d\theta$
$f’(\theta)=sin(\theta)+C_1$
Substitute in $f’(\frac{\pi}{2})=1$ to solve for $C_1$
$1=sin(\frac{\pi}{2})+C_1$
$C_1=0$
$f’(\theta)=sin(\theta)$
$f(\theta)=\int sin(\theta)d\theta$
$f(\theta)=-cos(\theta)+C_2$
Substitute in $f(\frac{\pi}{2})=6$ to solve for $C_2$
$6=-cos(\frac{\pi}{2})+C_2$
$C_2=6$
$f(\theta)=-cos(\theta)+6$
