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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?

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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 $\dfrac{\sin\left(\star\right)}{\cos \left(\star\right)} = \tan \left(\star\right)$. $$\begin{array}{rcl} \dfrac{b}{a} &=& -\color{green}{\dfrac{\sin \dfrac{\alpha - \beta}{2}}{\cos \dfrac{\alpha - \beta}{2}}}\\ &=& - \color{green} {\tan \dfrac{\alpha - \beta}{2}}\\ -\dfrac{b}{a} &=& \tan \dfrac{\alpha - \beta}{2} \end{array}$$

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  • $\begingroup$ That is an extremely well written answer. Thanks! $\endgroup$ – Aditya Sep 14 at 16:13
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Since $\frac{a}{2}=\sin\frac{\alpha+\beta}{2}\cos\frac{\alpha-\beta}{2}$ while $-\frac{b}{2}=\sin\frac{\alpha+\beta}{2}\sin\frac{\alpha-\beta}{2}$, $\tan\frac{\alpha-\beta}{2}=-\frac{b}{a}$.

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