If $sin x + sin y = a$ and $cos x + c o s y = b$ show that
$sin (x + y) = 2 a b / (a^{2} + b^{2})$

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Solution

Dividing the given relation We get
$\frac{sin x + sin y}{c o s x + cos y} = \frac{a}{b}$
or $\frac{2 sin [(x + y) / 2] \cdot cos [(x - y) / 2]}{2 cos [(x + y) / 2] \cdot cos [(x - y) / 2]} = \frac{a}{b}$
$∴ tan (\frac{x + y}{2}) = \frac{a}{b}$
Now $sin A$ = $\frac{2 tan (A / 2)}{1 + {tan}^{2} (A / 2)}$
$∴ sin (x + y) = \frac{2 \cdot a / b}{1 + a^{2} / b^{2}} = \frac{2 a b}{a^{2} + b^{2}}$

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