Answer
$$g''(\pi/6) =\sqrt{3}-\frac{\pi }{12} $$
Work Step by Step
Given $$g(\theta)=\theta\sin \theta$$
Since
\begin{align*}
g'(\theta)&=\theta\cos \theta+ \sin \theta\\
g''(\theta)&=-\theta\sin \theta+ 2\cos \theta
\end{align*}
then
$$g''(\pi/6) =-(\pi/6)\sin(\pi/6)+ 2\cos (\pi/6)=\sqrt{3}-\frac{\pi }{12} $$
