Viète's formula

Remarks

By induction, $a_n = 2 \cos(\pi / 2^{n+1})$, using $2\cos(\theta/2) = \sqrt{2 + 2\cos\theta}$ for $0 \le \theta \le \pi$. The double-angle formula $\sin 2t = 2 \sin t \cos t$, applied $N$ times, gives $$\sin x = 2^N \sin(x / 2^N) \prod_{n=1}^{N} \cos(x / 2^n).$$ For $x = \pi/2$ this says $\prod_{n=1}^{N} \cos(\pi / 2^{n+1}) = \frac{1}{2^N \sin(\pi / 2^{N+1})}$. Since $2^{N+1} \sin(\pi / 2^{N+1}) \to \pi$, the right side tends to $2/\pi$. The formula appeared in work of François Viète published in 1593. It converges fast: each factor adds about $0.6$ correct decimal digits of $\pi$.

Going deeper

For a student

The proof uses only the sine and cosine double-angle formula and the limit $\sin(t)/t \to 1$. The same computation gives the product $\frac{\sin x}{x} = \prod_{n \ge 1} \cos(x / 2^n)$ for every $x \ne 0$. Viète's formula is one of the first infinite products in mathematics, and the Wallis product is another classical product for $\pi$.