Product To Sum Formulas - Practice Questions & MCQ

Sum/Difference into Product-

1. $\sin \alpha+\sin \beta=2 \sin \left(\frac{\alpha+\beta}{2}\right) \cos \left(\frac{\alpha-\beta}{2}\right)$
2. $\sin \alpha-\sin \beta=2 \sin \left(\frac{\alpha-\beta}{2}\right) \cos \left(\frac{\alpha+\beta}{2}\right)$
3. $\cos \alpha-\cos \beta=-2 \sin \left(\frac{\alpha+\beta}{2}\right) \sin \left(\frac{\alpha-\beta}{2}\right)$
4. $\cos \alpha+\cos \beta=2 \cos \left(\frac{\alpha+\beta}{2}\right) \cos \left(\frac{\alpha-\beta}{2}\right)$

Proof

These formulas can be derived from the product-to-sum identities. For example, with a few substitutions, we can derive the sum-to-product identity for sine.
Let : $\frac{u+v}{2}=\alpha$ and $\frac{u-v}{2}=\beta$
Then,

$$
\begin{aligned}
\alpha+\beta & =\frac{u+v}{2}+\frac{u-v}{2} \\
& =\frac{2 u}{2} \\
& =u
\end{aligned}
$$

$$
\begin{aligned}
\alpha-\beta & =\frac{u+v}{2}-\frac{u-v}{2} \\
& =\frac{2 v}{2} \\
& =v
\end{aligned}
$$

Thus, by replacing $\alpha$ and $\beta$ in the product-to-sum formula with the substitute expressions, we have

$$
\begin{aligned}
2 \sin \alpha \cos \beta & =\sin (\alpha+\beta)+\sin (\alpha-\beta) \\
2 \sin \left(\frac{u+v}{2}\right) \cos \left(\frac{u-v}{2}\right) & =\sin u+\sin v, \quad \text { Substitute for }(\alpha+\beta) \text { and }(\alpha-\beta)
\end{aligned}
$$

The other sum-to-product identities are derived similarly.