## Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

$\dfrac{1}{2} [\sin (2 \theta) - \sin (\theta) ]$
Recall the Product to Sum Identities: $a) \sin x \cos y =\dfrac{1}{2} [\sin (x-y) +\sin (x+y)] \\ b) \sin x \sin y =\dfrac{1}{2} [\cos (x-y) -\cos (x+y)] \\ c) \cos x \cos y =\dfrac{1}{2} [\cos (x-y) +\cos (x+y)]$ By identity $(a)$, we have: $\sin ( \dfrac{3\theta}{2}) \ \cos (\dfrac{ \theta}{2})=\dfrac{1}{2} [\sin [( \dfrac{3\theta}{2}) + ( \dfrac{\theta}{2}) ]+\sin[( \dfrac{3\theta}{2}) - ( \dfrac{\theta}{2}) ] ] \\=\dfrac{1}{2} [\sin (2 \ \theta) - \sin (\theta) ]$