If sin α + sin β=a and cos α+cos β=b,

Solution:

The given equations are $\sin \alpha+\sin \beta=a$ and $\cos \alpha+\cos \beta=b$.

(i)

$\because \sin C+\sin D=2 \sin \frac{C+D}{2} \cos \frac{C-D}{2}$

$\therefore 2 \sin \frac{\alpha+\beta}{2} \cos \frac{\alpha-\beta}{2}=a \quad \ldots(1)$

Now, using the identity $\sin C+\sin D=2 \sin \frac{C+D}{2} \cos \frac{C-D}{2}$ for the LHS of $\cos \alpha+\cos \beta=b$, we get

$2 \cos \frac{\alpha+\beta}{2} \cos \frac{\alpha-\beta}{2}=b$         ....(2)

On dividing (1) by (2), we get

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

We know,

$\sin \theta=\frac{2 \tan \frac{\theta}{2}}{1+\tan ^{2} \frac{\theta}{2}}$

$\therefore \sin (\alpha+\beta)=\frac{2 \tan \left(\frac{\alpha+\beta}{2}\right)}{1+\tan ^{2}\left(\frac{\alpha+\beta}{2}\right)}$

$\Rightarrow \sin (\alpha+\beta)=\frac{2 \times \frac{a}{b}}{1+\frac{a^{2}}{2}}=\frac{2 a b}{a^{2}+b^{2}}$

(ii) 

On squaring $\sin \alpha+\sin \beta=a$ and $\cos \alpha+\cos \beta=b$ and adding them, we get

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

$\Rightarrow 1+1+2(\sin \alpha \sin \beta+\cos \alpha \cos \beta)=a^{2}+b^{2}$

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

$\Rightarrow 2 \cos (\alpha-\beta)=a^{2}+b^{2}-2 \quad(\because \cos (\mathrm{A}-\mathrm{B})=\sin \mathrm{A} \sin \mathrm{B}+\cos \mathrm{A} \cos \mathrm{B})$

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