Archimedes and the half angle

Archimedes found $\pi$ by drawing polygons inside a circle and doubling the number of sides again and again. A regular polygon with $n$ sides inside a circle of radius $1$ has half-perimeter $n\sin\dfrac{\pi}{n}$, which comes closer and closer to $\pi$. He had no calculator, but he had the half-angle formulae, and so do you.

The idea. From $c = \cos\dfrac{2\pi}{n}$, the half angle gives $\sin\dfrac{\pi}{n} = \sqrt{\dfrac{1 - c}{2}}$ and $\cos\dfrac{\pi}{n} = \sqrt{\dfrac{1 + c}{2}}$. The second is the $c$ for twice as many sides. You start from a square, where $c = \cos\dfrac{\pi}{2} = 0$.

Your tasks.

1. Make half_cos(c) return $\sqrt{\dfrac{1 + c}{2}}$, the cosine of half the angle.
2. In the loop, set s to $\sin\dfrac{\pi}{n}$ from c, with the half-angle formula.
3. Set sides_needed to the first n in ns whose estimate is within $0.001$ of $\pi$.