If + =a and + =b, show that ( + )= 2ab a^2+b^2

Question

If $\sin\alpha+\sin\beta=\text{a}$ and $\cos\alpha+\cos\beta=\text{b},$ show that
$\sin(\alpha+\beta)=\frac{2\text{ab}}{\text{a}^2+\text{b}^2}$

✓

Answer

$\text{a}^2+\text{b}^2=(\sin\alpha+\sin\beta)^2+(\cos\alpha+\cos\beta)^2$
$\Rightarrow\text{a}^2+\text{b}^2=\sin^2\alpha+\sin^2\beta+2\sin^2\alpha\sin^2\beta+\cos^2\alpha+\cos^2\beta+2\cos\alpha\cos\beta$
$\Rightarrow\text{a}^2+\text{b}^2=\sin^2\alpha+\cos^2\alpha+\sin^2\beta+\cos^2\beta+2(\sin\alpha+\sin\beta+\cos\alpha\cos\beta)$
$\Rightarrow\text{a}^2+\text{b}^2=2+2\cos(\alpha-\beta)\cdots(1)$
Now,
$\text{b}^2-\text{a}^2=(\cos\alpha+\cos\beta)^2-(\sin\alpha-\sin\beta)^2$
$\Rightarrow\text{b}^2-\text{a}^2=\cos^2+\cos^2\beta+\sin^2\alpha-\sin^2\beta+2\cos\alpha+\cos\beta-2\sin\alpha\sin\beta$
$\Rightarrow\text{b}^2-\text{a}^2=(\cos^2\alpha-\sin^2\beta)+(\cos^2\beta-\sin^2\alpha)-2\cos(\alpha+\beta)$
$\Rightarrow\text{b}^2-\text{a}^2=2\cos(\alpha+\beta)\cos(\alpha-\beta)+2\cos(\alpha-\beta)$
$\Rightarrow\text{b}^2-\text{a}^2=\cos(\alpha+\beta)\Big(2+2\cos(\alpha-\beta)\Big)\cdots(2)$
From (1) and (2), we have
$\text{b}^2-\text{a}^2=\cos(\alpha+\beta)(\text{a}^2+\text{b}^2)$
$\Rightarrow\frac{\text{b}^2-\text{a}^2}{\text{a}^2-\text{b}^2}=\cos(\alpha+\beta)$
$\Rightarrow\sin(\alpha+\beta)=\sqrt{1-\cos^2(\alpha+\beta)}$
$\Rightarrow\sin(\alpha+\beta)=\sqrt{1-\Big(\frac{\text{b}^2-\text{a}^2}{\text{b}^2+\text{a}^2}\Big)^2}=\sqrt{\frac{\text{b}^4+\text{a}^4-\text{b}^4-\text{a}^4+4\text{a}^2\text{b}^2}{(\text{b}^2+\text{a}^2)}}$
$\Rightarrow\sin(\alpha+\beta)=\frac{2\text{ab}}{\text{a}^2+\text{b}^2}$