Question

express each product as a sum containing only sines or only cosines

$\sin \vartheta \sin 2\vartheta$

Short answer

$\frac{1}{2}[\cos \vartheta -\cos 3\vartheta ]$

Step-by-step solution

Step 1. Given expression 

$\sin \vartheta \sin 2\vartheta$

Step 2. expressing them in terms of single sine or cosine

Using identity :

$\cos x-\cos y=-2\sin \left(\frac{x+y}{2}\right)\sin \left(\frac{x-y}{2}\right)$

We get :

$$\begin{gathered} \sin \vartheta \sin 2\vartheta =\frac{1}{2}\left[\cos \right(2\vartheta -\vartheta )-\cos (2\vartheta +\vartheta \left)\right] \\ \sin \vartheta \sin 2\vartheta =\frac{1}{2}[\cos \vartheta -\cos 3\vartheta ] \end{gathered}$$