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  "body": "\n## Proof\n\n1. Fix a real $x$. Let $v_n = \\frac{(-1)^n x^{2n}}{(2n)!}$ for $n \\ge 0$, where $k!$ is the [[factorial|0WFCGVMRRC]].\n2. If $x = 0$, then $v_0 = 1$ and $v_n = 0$ for $n \\ge 1$, so $\\sum \\abs{v_n}$ converges.\n3. If $x \\ne 0$, then $v_n \\ne 0$ and $\\frac{\\abs{v_{n+1}}}{\\abs{v_n}} = \\frac{x^2}{(2n+1)(2n+2)} \\to 0 \u003c 1$.\n4. By the [[ratio test|0QNY6ZBXPN]], $\\sum v_n$ converges [[absolutely|0W3K3QPEWT]] in this case too.\n5. By the [[definition of cosine|0WB57D9488]], $\\cos x = \\sum_{n=0}^{\\infty} v_n = 1 - \\frac{x^2}{2!} + \\frac{x^4}{4!} - \\cdots$.\n6. It remains to identify this series as the Maclaurin series. By the [[derivative of sine|00R4MWW532]] and the [[derivative of cosine|0V2DV29EGT]], $\\sin' = \\cos$ and $\\cos' = -\\sin$ on $\\mathbb{R}$.\n7. By induction on $k$, the [[higher derivatives|084NGSTWM7]] exist for all orders: $\\cos^{(4k)} = \\cos$, $\\cos^{(4k+1)} = -\\sin$, $\\cos^{(4k+2)} = -\\cos$ and $\\cos^{(4k+3)} = \\sin$. So $\\cos$ is smooth.\n8. In the defining series at $0$, only the first term of the cosine series is nonzero, and every term of the sine series is zero. So $\\cos 0 = 1$ and $\\sin 0 = 0$.\n9. By steps 7 and 8, $\\cos^{(m)}(0) = 0$ for odd $m$, $\\cos^{(m)}(0) = 1$ for $m = 4k$, and $\\cos^{(m)}(0) = -1$ for $m = 4k + 2$.\n10. So for $m = 2n$ we get $\\cos^{(m)}(0) = (-1)^n$, since $n$ is even exactly when $m = 4k$.\n11. The [[Maclaurin series|043SJ5RF59]] $\\sum_m \\frac{\\cos^{(m)}(0)}{m!} x^m$ thus has zero terms for odd $m$, and the term $\\frac{(-1)^n x^{2n}}{(2n)!}$ for $m = 2n$.\n12. Removing the zero terms, the Maclaurin series of $\\cos$ is $\\sum_{n=0}^{\\infty} \\frac{(-1)^n x^{2n}}{(2n)!}$, the series of the statement.\n\n## Idea\n\nThe terms shrink like $\\frac{1}{(2n)!}$, which beats any power of $x$, so the ratio test gives absolute convergence. With cosine defined by this series, the equality is the definition. The derivatives of $\\cos$ at $0$ cycle through $1, 0, -1, 0$, and these values give back the same coefficients.\n",
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