Question

$\cos \alpha +\cos \beta =a,\sin \alpha +\sin \beta =b$

Answer 1

$\cos \alpha +\cos \beta =a......\left(1\right)$

$\Rightarrow a=2\cos \frac{\alpha +\beta }{2}\cos \frac{\alpha -\beta }{2}....\left(2\right)$

$\sin \alpha +\sin \beta =b.....\left(3\right)$

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

Dividing $\left(4\right)$ by $\left(2\right)$

$\tan \frac{\alpha +\beta }{2}=\frac{b}{a}$

A) $\cos (\alpha +\beta )=\frac{1-\left(b^2/a^2\right)}{1+(b^2+a^2)}=\frac{a^2-b^2}{a^2+b^2}$

B) $\sin (\alpha +\beta )=\frac{2\left(b/a\right)}{1+\left(b^2/a^2\right)}=\frac{2ab}{a^2+b^2}$

C) Squaring and adding $\left(1\right)$ and $\left(3\right)$, we get

$a^2+b^2={\sin }^2\alpha +{\sin }^2\beta +{\cos }^2\alpha +{\cos }^2\beta +2\cos \alpha \cos \beta +2\sin \alpha \sin \beta$

$\Rightarrow a^2+b^2-2=2\cos (\alpha -\beta )\Rightarrow \cos (\alpha -\beta )=\frac{a^2+b^2-2}{2}$

D) $\tan \frac{\alpha +\beta }{2}=\frac{b}{a}$