Question

If \[\sin \alpha + \sin \beta = a \text{ and }  \cos \alpha + \cos \beta = b\] , prove that 
(i)\[\sin \left( \alpha + \beta \right) = \frac{2ab}{a^2 + b^2}\]

Answer 1

The given equations are \[\sin \alpha + \sin \beta = a \text{ and }  \cos \alpha + \cos \beta = b\]

(i) \[\because \text{ sin } C + \text{ sin } D = 2\sin\frac{C + D}{2}\cos\frac{C - D}{2}\]
\[ \therefore 2\sin\frac{\alpha + \beta}{2}\cos\frac{\alpha - \beta}{2} = a . . . (1)\] 

Now, using the identity

\[\text{ sin } C + \text{ sin } D = 2\sin\frac{C + D}{2}\cos\frac{C - D}{2}\]  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,

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

\[\therefore \sin\left( \alpha + \beta \right) = \frac{2\tan\left( \frac{\alpha + \beta}{2} \right)}{1 + \tan^2 \left( \frac{\alpha + \beta}{2} \right)}\]
\[ \Rightarrow \sin\left( \alpha + \beta \right) = \frac{2 \times \frac{a}{b}}{1 + \frac{a^2}{b^2}} = \frac{2ab}{a^2 + b^2}\]