Answer
$$f'\left( \theta \right) = \theta \cos \theta - {\cos ^2}\theta + \sin \theta + {\sin ^2}\theta $$
Work Step by Step
$$\eqalign{
& f\left( \theta \right) = \left( {\theta - \cos \theta } \right)\sin \theta \cr
& {\text{Differentiating}} \cr
& f'\left( \theta \right) = \frac{d}{{d\theta }}\left[ {\left( {\theta - \cos \theta } \right)\sin \theta } \right] \cr
& {\text{Use the product rule for derivatives }}\left( {hg} \right)' = hg' + gh' \cr
& {\text{Let }}h = \left( {\theta - \cos \theta } \right){\text{ and }}g = \sin \theta ,{\text{ then}} \cr
& f'\left( \theta \right) = \left( {\theta - \cos \theta } \right)\frac{d}{{d\theta }}\left[ {\sin \theta } \right] + \sin \theta \frac{d}{{d\theta }}\left[ {\theta - \cos \theta } \right] \cr
& {\text{Use the Derivatives of Trigonometric Functions }} \cr
& f'\left( \theta \right) = \left( {\theta - \cos \theta } \right)\left( {\cos \theta } \right) + \sin \theta \left[ {1 - \left( { - \sin \theta } \right)} \right] \cr
& {\text{Simplify}} \cr
& f'\left( \theta \right) = \left( {\theta - \cos \theta } \right)\left( {\cos \theta } \right) + \sin \theta \left( {1 + \sin \theta } \right) \cr
& f'\left( \theta \right) = \theta \cos \theta - {\cos ^2}\theta + \sin \theta + {\sin ^2}\theta \cr} $$
