# Proving that $\frac{\sin \alpha + \sin \beta}{\cos \alpha + \cos \beta} =\tan \left ( \frac{\alpha+\beta}{2} \right )$

Using double angle identities a total of four times, one for each expression in the left hand side, I acquired this.

$$\frac{\sin \alpha + \sin \beta}{\cos \alpha + \cos \beta} = \frac{\sin \left ( \frac{\alpha}{2}\right ) \cos \left ( \frac{\alpha}{2}\right ) + \sin \left ( \frac{\beta}{2}\right ) \cos \left ( \frac{\beta}{2}\right )}{\cos^2 \left ( \frac{\alpha}{2} \right) - \sin ^2 \left ( \frac{\beta}{2} \right )}$$

But I know that if $\alpha$ and $\beta$ are angles in a triangle, then this expression should simplify to

$$\tan \left ( \frac{\alpha + \beta}{2} \right )$$

I can see that the denominator becomes $$\cos \left ( \frac{\alpha + \beta}{2} \right )$$

But I cannot see how the numerator becomes

$$\sin \left ( \frac{\alpha + \beta}{2} \right )$$

What have I done wrong here?

• Oct 5, 2015 at 10:46
• Can you elaborate? Oct 5, 2015 at 10:49
• How can you see that the denominator becomes $\cos \left ( \frac{\alpha + \beta}{2} \right )$? Oct 5, 2015 at 10:50

$$\sin\alpha + \sin\beta = 2\sin(\frac{\alpha+\beta}{2})\cos(\frac{\alpha-\beta}{2}).$$

$$\cos\alpha + \cos\beta = 2\cos(\frac{\alpha+\beta}{2})\cos(\frac{\alpha-\beta}{2}).$$

So, you get the conclusion.

Another approach:

Put: $\tan (\alpha/2)=a$ and $\tan (\beta/2)=b$.

Than: $$\sin \alpha= \dfrac{2a}{1+a^2} \qquad \cos \alpha=\dfrac{1-a^2}{1+a^2}$$ $$\sin \beta= \dfrac{2b}{1+b^2} \qquad \cos \beta=\dfrac{1-b^2}{1+b^2}$$

and: $$\frac{\sin \alpha + \sin \beta}{\cos \alpha + \cos \beta} = \dfrac{2a(1+b^2)+2b(1+a^2)}{(1-a^2)(1+b^2)+(1-b^2)(1+a^2)}$$ that, after a bit of algebra, becomes: $$=\dfrac{a+b}{1-ab}=\dfrac{\tan \alpha/2+\tan \beta/2}{1-\tan \alpha/2 \tan \beta/2}= \tan \left(\dfrac{\alpha +\beta}{2} \right)$$

Assume that $$P=(\cos\ a,\sin\ b),\ Q=(\cos\ b,\sin\ b)$$ Then what is slope of line passing through $$-Q,\ P$$ :

For convenience $$0, let $$R=(0,1)$$. Then $$\angle\ ROQ=\frac{\pi}{2}-b,\ O=(0,0)$$

and $$\angle\ Q(-Q)P = \frac{b-a}{2}$$

When $$\angle\ PP'(-Q) = \pi/2$$, then $$\angle\ P(-Q)P'= \frac{a+b}{2}$$

Hence $$\tan\ \frac{a+b}{2} = \frac{{\rm difference \ of}\ y-{\rm coordinates\ of}\ P,\ -Q }{{\rm difference \ of}\ x-{\rm coordinates \ of}\ P,\ -Q }$$