If $\sin\alpha+\sin\beta=a$ and $\cos\alpha-\cos\beta=b$, then what is $\tan(\frac{\alpha-\beta}{2})$? If I square both the equations 
$$2+2\sin(\alpha-\beta)=a^2+b^2$$
$$\sin(\alpha-\beta)=\frac{a^2+b^2-2}{2}$$
Since $\sin2\theta=\frac{2\tan\theta}{1+\tan^2\theta}$, then $$\sin(\alpha-\beta)=\frac{2\tan\frac{\alpha-\beta}{2}}{1+\left(\tan\frac{\alpha-\beta}{2}\right)^2}$$
It’s obviously the too long to solve, so is there a shorter way to do this?
 A: Since $\frac{a}{2}=\sin\frac{\alpha+\beta}{2}\cos\frac{\alpha-\beta}{2}$ while $-\frac{b}{2}=\sin\frac{\alpha+\beta}{2}\sin\frac{\alpha-\beta}{2}$,  $\tan\frac{\alpha-\beta}{2}=-\frac{b}{a}$.
A: Denote $x = \dfrac{\alpha + \beta}{2}$ and $y = \dfrac{\alpha - \beta}{2}$.
Question: Are the any relations between $x, y$ and $\alpha, \beta$?
Answer:

 Yes. $$x - y = \dfrac{\alpha + \beta}{2} - \dfrac{\alpha - \beta}{2} = \dfrac{2\beta}{2} = \beta\\x + y = \dfrac{\alpha + \beta}{2} + \dfrac{\alpha - \beta}{2} = \dfrac{2\alpha}{2} = \alpha$$ Therefore, whenever we see $\alpha$ and $\beta$, we can substitute $x + y$ and $x - y$ respectively.

We try to express $a$ and $b$ in terms of trigonometric functions of $x$ and $y$.
We have:
$$\begin{array}{rcl}
a &=& \sin \alpha + \sin \beta\\
&=& \sin \left(x + y\right) + \sin (x - y)\\
&=& (\sin x \cos y + \cos x \sin y) + (\sin x \cos y - \cos x \sin y)\\
&=& 2 \sin x \cos y\\
&=& 2 \sin \dfrac{\alpha + \beta}{2} \cos \dfrac{\alpha - \beta}{2}
\end{array}$$
Also we have:
$$\begin{array}{rcl}
b &=& \cos \alpha - \cos \beta\\
&=& \cos(x + y) - \cos(x - y)\\
&=& (\cos x \cos y - \sin x \sin y) - (\cos x \cos y + \sin x \sin y)\\
&=& -2\sin x \sin y\\
&=& -2 \sin \dfrac{\alpha + \beta}{2} \sin \dfrac{\alpha - \beta}{2}
\end{array}$$
We consider $\dfrac{b}{a}$. (Why?)
$$\begin{array}{rcl}
\dfrac{b}{a} &=& \dfrac{-\color{red}{2\sin \dfrac{\alpha + \beta}{2}} \sin \dfrac{\alpha - \beta}{2}}{ \color{red}{2\sin \dfrac{\alpha + \beta}{2}} \cos \dfrac{\alpha - \beta}{2}}\\
&=& - \dfrac{\sin \dfrac{\alpha - \beta}{2}}{\cos \dfrac{\alpha - \beta}{2}}
\end{array}$$
Recall that $\dfrac{\sin\left(\star\right)}{\cos \left(\star\right)} = \tan \left(\star\right)$.
$$\begin{array}{rcl}
\dfrac{b}{a} &=& -\color{green}{\dfrac{\sin \dfrac{\alpha - \beta}{2}}{\cos \dfrac{\alpha - \beta}{2}}}\\
&=& - \color{green} {\tan \dfrac{\alpha - \beta}{2}}\\
-\dfrac{b}{a} &=& \tan \dfrac{\alpha - \beta}{2}
\end{array}$$
