Additions of sines.

Question

What would these expressions evaluate to and why? A proof would be appreciated.

$$\sin(A) ± \sin(B)$$ $$\cos(A) ± \cos(B)$$

If there’s anything else I should know, please inform me. Thank you.

Answer 1

If you solve $\left\{\begin{array}{c} A = u+v \\ B = u-v \end{array}\right\}$ for $u$ and $v$, you get $\left\{ \begin{array}{c} u = \frac{A+B}{2} \\ v = \frac{A-B}{2} \end{array} \right\}$.

So \begin{align} \sin A + \sin B &= \sin(u+v) + \sin(u-v) \\ &= (\sin u \ \cos v + \cos u \ \sin v) + (\sin u \ \cos v - \cos u \ \sin v) \\ &= 2 \sin u \ \cos v \\ &= 2 \sin \frac{A+B}{2} \ \cos \frac{A-B}{2} \\ \end{align}

The other formulas are computed similarly.