Sum-to-Product and Product-to-Sum Formulas - Practice Questions & MCQ

Product into Sum/Difference
1. $2 \cos \alpha \cos \beta=[\cos (\alpha-\beta)+\cos (\alpha+\beta)]$
2. $2 \sin \alpha \sin \beta=[\cos (\alpha-\beta)-\cos (\alpha+\beta)]$
3. $2 \sin \alpha \cos \beta=[\sin (\alpha+\beta)+\sin (\alpha-\beta)]$
4. $2 \cos \alpha \sin \beta=[\sin (\alpha+\beta)-\sin (\alpha-\beta)]$

Proof:
We can derive the product-to-sum formula from the sum and difference identities.

$$
\begin{gathered}
\cos \alpha \cos \beta+\sin \alpha \sin \beta=\cos (\alpha-\beta) \\
+\cos \alpha \cos \beta-\sin \alpha \sin \beta=\cos (\alpha+\beta) \\
\hline 2 \cos \alpha \cos \beta=\cos (\alpha-\beta)+\cos (\alpha+\beta) \\
\begin{array}{c}
\sin (\alpha+\beta)=\sin (\alpha) \cos (\beta)+\cos (\alpha) \sin (\beta) \\
+\quad \sin (\alpha-\beta)=\sin (\alpha) \cos (\beta)-\cos (\alpha) \sin (\beta) \\
\hline
\end{array}+\frac{\sin (\alpha+\beta)+\sin (\alpha-\beta)=2 \sin (\alpha) \cos (\beta)}{\cos (\alpha-\beta)=\cos \alpha \cos \beta+\sin \alpha \sin \beta} \\
-\cos (\alpha+\beta)=-(\cos \alpha \cos \beta-\sin \alpha \sin \beta) \\
\hline \cos (\alpha-\beta)-\cos (\alpha+\beta)=2 \sin \alpha \sin \beta
\end{gathered}
$$