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  "body": "\n## Statement\n\nLet $a_1 = \\sqrt{2}$ and $a_{n+1} = \\sqrt{2 + a_n}$ for $n \\ge 1$.\nThen the infinite product $\\prod_{n=1}^{\\infty} a_n / 2$ [[converges|H1Z2TKY3EQ]], and\n$$\\prod_{n=1}^{\\infty} \\frac{a_n}{2} = \\frac{\\sqrt2}{2} \\cdot \\frac{\\sqrt{2 + \\sqrt2}}{2} \\cdot \\frac{\\sqrt{2 + \\sqrt{2 + \\sqrt2}}}{2} \\cdots = \\frac{2}{\\pi},$$\nwhere $\\pi$ is the number [[pi|0C1VYVNV2N]].\n\n## Remarks\n\nBy induction, $a_n = 2 \\cos(\\pi / 2^{n+1})$, using $2\\cos(\\theta/2) = \\sqrt{2 + 2\\cos\\theta}$ for $0 \\le \\theta \\le \\pi$.\nThe double-angle formula $\\sin 2t = 2 \\sin t \\cos t$, applied $N$ times, gives\n$$\\sin x = 2^N \\sin(x / 2^N) \\prod_{n=1}^{N} \\cos(x / 2^n).$$\nFor $x = \\pi/2$ this says $\\prod_{n=1}^{N} \\cos(\\pi / 2^{n+1}) = \\frac{1}{2^N \\sin(\\pi / 2^{N+1})}$.\nSince $2^{N+1} \\sin(\\pi / 2^{N+1}) \\to \\pi$, the right side tends to $2/\\pi$.\nThe formula appeared in work of François Viète published in 1593.\nIt converges fast: each factor adds about $0.6$ correct decimal digits of $\\pi$.\n\n## Explanations\n\n### L1 human\n\nViète's formula writes $2/\\pi$ as an infinite product of nested square roots of $2$.\nIt comes from inscribing regular polygons with $4, 8, 16, \\ldots$ sides in a circle and comparing their areas.\nAs the number of sides doubles, the polygons fill the circle, and the ratio of areas gives the next factor.\n\n### L2 human\n\nThe proof uses only the [[sine and cosine|08WFZP6P89]] double-angle formula and the limit $\\sin(t)/t \\to 1$.\nThe same computation gives the product $\\frac{\\sin x}{x} = \\prod_{n \\ge 1} \\cos(x / 2^n)$ for every $x \\ne 0$.\nViète's formula is one of the first infinite products in mathematics, and the [[Wallis product|0XCYA4YERS]] is another classical product for $\\pi$.\n",
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