{
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  "body": "\n## Statement\n\nFor each $z \\in \\mathbb{C}$, the series $\\sum_{n \\ge 0} (-1)^n z^{2n} / (2n)!$ converges absolutely, and\n$$\\cos z = \\sum_{n=0}^{\\infty} (-1)^n \\frac{z^{2n}}{(2n)!} = 1 - \\frac{z^2}{2!} + \\frac{z^4}{4!} - \\cdots,$$\nwhere $\\cos$ is the [[complex cosine|0A7YDKRNR6]].\n\n## Remarks\n\nProof: put $iz$ and $-iz$ in the series of the [[complex exponential|054VXHMCA3]] and add. The terms with odd powers cancel, and $i^{2n} = (-1)^n$.\nIn the same way, $\\sin z = \\sum_{n \\ge 0} (-1)^n z^{2n+1} / (2n+1)!$ for each $z \\in \\mathbb{C}$.\nThe series shows that $\\cos$ is an even function: $\\cos(-z) = \\cos z$.\n\n## Explanations\n\n### L1 human\n\nThe series of $\\cos$ has only the even powers of $e^z$'s series, with signs that alternate.\nFor example, $\\cos(0.1) \\approx 1 - 0.005 = 0.995$ from the first two terms.\n",
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  "msc": [
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  "name": "Power series of complex cosine",
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