Proving $\tan\left( \frac {a+b}{2}\right) = \frac{\sin a+\sin b} {\cos a + \cos b }$

Prove $$\tan\left( \frac {a+b}{2}\right) = \frac{\sin a+\sin b} {\cos a + \cos b }$$

Can someone help? I separated $\tan$ and did double-angle, but I just went into a circle and couldn't get the trig functions without the halves.

2 Answers

Use the formulas $$\sin a+\sin b=2\sin\left(\frac{a+b}{2}\right)\cos\left(\frac{a-b}{2}\right),$$ $$\cos a+\cos b=2\cos\left(\frac{a+b}{2}\right)\cos\left(\frac{a-b}{2}\right).$$

Therefore, we have

$$\frac{\sin a+\sin b}{\cos a+\cos b}=\frac{\sin\displaystyle\left(\frac{a+b}{2}\right)}{\cos\displaystyle\left(\frac{a+b}{2}\right)}=\tan\left(\frac{a+b}{2}\right).$$

If $$a,b$$ are acute angles one can proceed as follows:

Draw a parallelogram with all four sides of length $$1,$$ as follows. The first side has one endpoint at $$(0,0)$$ and the angle between it and the $$x$$-axis is $$a$$. Its other endpoint is therefore at $$(\cos a,\sin a).$$ The second side begins at that latter point and its angle with the $$x$$-axis is $$b.$$ Its other endpoint is therefore $$(\sin a + \sin b, \cos a+\cos b).$$ The tangent of the angle from the origin to this last point is therefore $$(\sin a+\sin b)/(\cos a + \cos b).$$ But the geometric symmetry of the parallelogram shows that that angle is $$(a+b)/2.$$