<s>[INST] Derive Formula for Sine Half Angle I got stack with another problem (From the book: The Forgotten Art of Spherical Trigonometry):

Not having OC as 1 always creates me complicated formulas to define the rest. Once again looking for the smart track that I am missing.
 [/INST] For ease of notation, let $\beta = \alpha/2$.  Note that in the figure, $\triangle ABO \cong \triangle CBO$.  Hence $$AO = CO = \cos \beta, \quad AB = CB = \sin \beta.$$  Since $DO/CO = \cos \alpha$, it follows that $DO = \cos \alpha \cos \beta$.  Similarly, $CD/CO = \sin \alpha$, hence $CD = \sin \alpha \cos \beta$.  But $ABED$ is a rectangle, thus $$BE = AD = AO - DO = \cos \beta - \cos \alpha \cos \beta = \cos \beta (1 - \cos \alpha),$$ and $$CE = CD - ED = CD - BA = \sin \alpha \cos \beta - \sin \beta.$$  Then by the Pythagorean theorem, $$\begin{align*} \sin^2 \beta &= BC^2 = BE^2 + CE^2 \\ &= \cos^2 \beta (1 - \cos \alpha)^2 + (\sin \alpha \cos \beta - \sin \beta)^2 \\ &= \cos^2 \beta (1 - \cos \alpha)^2 + \cos^2 \beta \sin^2 \alpha - 2 \sin \alpha \cos \beta \sin \beta + \sin^2 \beta, \end{align*}$$ and cancelling $\sin^2 \beta$ from both sides, rearranging, and factoring like terms gives $$\begin{align*} 2 \sin \alpha \cos \beta \sin \beta &= \cos^2 \beta \left( (1 - \cos \alpha)^2 + \sin^2 \alpha\right) \\ &= 2 \cos^2 \beta (1 - \cos \alpha).\end{align*}$$  Now cancelling again and isolating $\beta$ gives $$\cot \beta = \frac{\sin \alpha}{1 - \cos \alpha}.$$  All that is left now is to square both sides to get $$\cot^2 \beta = \frac{\sin^2 \alpha}{(1 - \cos \alpha)^2} = \frac{1 - \cos^2 \alpha}{(1 - \cos \alpha)^2} = \frac{1 + \cos \alpha}{1 - \cos \alpha},$$ so that $$\csc^2 \beta = 1 + \cot^2 \beta = \frac{2}{1 - \cos \alpha},$$ therefore, $\sin^2 \beta = \dfrac{1 - \cos \alpha}{2},$ or $$\sin \frac{\alpha}{2} = \sqrt{\frac{1 - \cos \alpha}{2}}.$$  Compared with Blue's diagram and answer, I think it should be fairly clear which approach is more elegant.
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