User:DoubleAW/Proof of tangent of an average



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