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.


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 $(a+b)/2.$


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