# Identity

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