If $ \sin \alpha + \sin \beta = a $ and $ \cos \alpha + \cos \beta = b $ , then show that $\sin(\alpha + \beta) = \frac {2ab } { a^2 + b^2} $ I've been able to do this, but I had to calculate $ \cos (\alpha + \beta) $ first. Is there a way to do this WITHOUT calculating $\cos(\alpha+\beta)$ first ?
Here's how I did it by calculating $\cos(\alpha+\beta)$ first
$ a^2 + b^2 = \sin ^2 \alpha + \sin ^2 \beta + 2 \sin \alpha \sin \beta + \cos ^2 \alpha + \cos ^2 \beta + 2 \cos \alpha \cos \beta $
$a^2 + b^2 = (\sin^2\alpha + \cos^2\alpha) + (\sin ^2 \beta + \cos^2 \beta) + 2(\cos\alpha\cos\beta + \sin\alpha\sin\beta)$
$a^2 + b^2 = 2 (1  + \cos(\alpha-\beta))$
$ \frac{a^2 + b^2}{2} = (1 + \cos(\alpha - \beta))$ 
$ b^2 - a^2 = (\cos ^2\alpha - \sin^2\alpha) + (\cos^2 \beta - \sin^2\beta) + 2\cos\alpha\cos\beta - 2\sin\alpha\sin\beta$
$b^2 - a^2 = (\cos^2\alpha - (1 - cos^2\alpha)) +(1-\sin^2\beta) - \sin^2\beta)) + 2(\cos\alpha\cos\beta - \sin\alpha\sin\beta) $
$b^2 - a^2 = 2 (\cos^2\alpha - \sin^2\beta + \cos(\alpha+\beta))$
$b^2 - a^2 = 2(\cos(\alpha+\beta)\cos(\alpha-\beta)+\cos(\alpha+\beta))$
$\frac{b^2 - a^2}{2} = \cos(\alpha+\beta)\{\cos(\alpha-\beta) + 1 \}$
$\frac{b^2 - a^2}{2} = \cos(\alpha+\beta)\{\frac{b^2+a^2}{2}\}$
$\cos(\alpha+\beta) = \frac {a^2 + b^2 } {a^2 - b^2}$
Then I just calculated $\sin(\alpha + \beta)$ by $1 - \cos^2(\alpha+\beta)$  

\begin{align*}
  b+ai &= e^{i\alpha}+e^{i\beta} \\[5pt]
  b-ai &= e^{-i\alpha}+e^{-i\beta} \\[5pt]
  \frac{b+ai}{b-ai} &=
  \frac{e^{i\alpha}+e^{i\beta}}{e^{-i\alpha}+e^{-i\beta}} \\[5pt]
  \frac{(b+ai)(b+ai)}{(b-ai)(b+ai)} &=
  \frac{e^{i(\alpha+\beta)}(e^{i\alpha}+e^{i\beta})}
       {e^{i(\alpha+\beta)}(e^{-i\alpha}+e^{-i\beta})} \\[5pt]
  \frac{(b+ai)^2}{b^2+a^2} &=
  \frac{e^{i(\alpha+\beta)}(e^{i\alpha}+e^{i\beta})}
       {e^{i\beta}+e^{i\alpha}} \\[5pt]
\frac{b^2-a^2}{a^2+b^2}+\frac{2ab}{a^2+b^2}i &=
  e^{i(\alpha+\beta)}
\end{align*}
The result follows by comparing the imaginary parts.
