User:DoubleAW/Proof of tangent of an average

O is the origin, A is the point (cos α, sin α) and B is the point (cos β, sin β)

Let $O$ be the center of a unit circle, $A$ and $B$ are points on the circle creating angles $\alpha$ and $\beta$ respectively, and $M$ is the midpoint of ${\overline {AB}}$.

$m\angle AOB=\alpha -\beta$, and ${\overline {OM}}$ is the angle bisector of $\angle AOB$ because the angle bisector to the non-congruent side of an isosceles triangle is always the perpendicular bisector to that side, meaning $m\angle AOM={\frac {\alpha -\beta }{2}}$. Subtracting that angle from $\alpha$ results in $m\angle MOQ=\alpha -{\frac {\alpha -\beta }{2}}={\frac {\alpha +\beta }{2}}$.

Because ${\overline {OM}}\perp {\overline {AB}}$, their slopes $m_{\overline {OM}}$ and $m_{\overline {AB}}$ are negative reciprocals of each other.

$m_{\overline {AB}}={\frac {\sin \beta -\sin \alpha }{\cos \beta -\cos \alpha }}$

$m_{\overline {OM}}=\tan \left({\frac {\alpha +\beta }{2}}\right)=-{\frac {\cos \beta -\cos \alpha }{\sin \beta -\sin \alpha }}\cdot {\frac {(\cos \beta +\cos \alpha )(\sin \alpha +\sin \beta )}{(\cos \alpha +\cos \beta )(\sin \beta +\sin \alpha )}}$

$m_{\overline {OM}}={\frac {-(\cos ^{2}\beta -\cos ^{2}\alpha )(\sin \alpha +\sin \beta )}{(\sin ^{2}\beta -\sin ^{2}\alpha )(\cos \alpha +\cos \beta )}}={\frac {\sin \alpha +\sin \beta }{\cos \alpha +\cos \beta }}$

Therefore, the tangent of the average of two angles $\alpha$ and $\beta$ is given as $\tan \left({\frac {\alpha +\beta }{2}}\right)={\frac {\sin \alpha +\sin \beta }{\cos \alpha +\cos \beta }}=-{\frac {\cos \beta -\cos \alpha }{\sin \beta -\sin \alpha }}$.