# 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?

• Tip: Your posts will look a lot better, and be easier to read, if your write $\sin$, \$\cos and so on. Sep 14 '19 at 14:45
• Sep 14 '19 at 15:08
• @saulspatz but it’s a real pain to do put a \ in front of everything, when your computer is broken and you have to type it on an iPad. (I get that you might not care about my personal problems, I just hope that you understand me predicament.) Sep 14 '19 at 15:18

Denote $$x = \dfrac{\alpha + \beta}{2}$$ and $$y = \dfrac{\alpha - \beta}{2}$$.

Question: Are the any relations between $$x, y$$ and $$\alpha, \beta$$?

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

• That is an extremely well written answer. Thanks! Sep 14 '19 at 16:13

Since $$\frac{a}{2}=\sin\frac{\alpha+\beta}{2}\cos\frac{\alpha-\beta}{2}$$ while $$-\frac{b}{2}=\sin\frac{\alpha+\beta}{2}\sin\frac{\alpha-\beta}{2}$$, $$\tan\frac{\alpha-\beta}{2}=-\frac{b}{a}$$.