While struggling with a Catronia Shearer problem I stumbled across a proof of the trig identity
which I hadn’t seen and probably wouldn’t have thought of if I’d been looking for it, though there’s nothing non-obvious about it in retrospect. Here it is. Draw a unit semicircle centered on O, with A and C the endpoints of the diameter. Draw chords AD and DC. (Angle ADC is a right angle.) Let angle CAD be . By the Inscribed Angle Theorem, angle COD is . Drop a perpendicular from D to the diameter at B.
Now and , so . But so , so . Therefore .