While struggling with a Catronia Shearer problem I stumbled across a proof of the trig identity

\cos^2\theta = \frac{1}{2}(1+\cos 2\theta)

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 \theta. By the Inscribed Angle Theorem, angle COD is 2\theta. Drop a perpendicular from D to the diameter at B.

Now \mathrm {AO} = 1 and \mathrm {OB} = \cos 2\theta, so \mathrm {AB} = 1+\cos 2\theta. But \mathrm {AC} = 2 so \mathrm {AD} = 2 \cos \theta, so \mathrm {AB} = 2 \cos^2 \theta. Therefore \cos^2\theta = \frac{1}{2}(1+\cos 2\theta).

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