#### Answer

$$ f''(\theta)
=2\cos \theta-\theta \sin \theta.$$

#### Work Step by Step

Recall the product rule: $(uv)'=u'v+uv'$
Recall that $(\sin x)'=\cos x$.
Recall that $(\cos x)'=-\sin x$.
Since $ f(\theta)=\theta \sin \theta $, then using the product rule we have
$$ f'(\theta)= (\theta)' \sin \theta +\theta (\sin \theta )'\\
=\sin \theta + \theta \cos \theta,$$ and again using the product rule, we have
$$ f''(\theta)=\cos \theta+(\theta)' \cos \theta+\theta (\cos \theta)'
\\= \cos \theta + \cos \theta-\theta \sin \theta\\
=2\cos \theta-\theta \sin \theta.$$