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  "body": "\n## Proof\n\n1. By the [[definition of cosine|0WB57D9488]], $\\cos x = \\sum_{m=0}^{\\infty} d_m x^m$, where $d_{2n} = \\frac{(-1)^n}{(2n)!}$ and $d_{2n+1} = 0$ for $n \\ge 0$. Here $k!$ is the [[factorial|0WFCGVMRRC]].\n2. Let $r \u003e 0$, and let $t_n = \\frac{r^{2n}}{(2n)!} \u003e 0$. Then $\\frac{t_{n+1}}{t_n} = \\frac{r^2}{(2n+1)(2n+2)} \\to 0$.\n3. By the [[ratio test|0QNY6ZBXPN]], $\\sum_n t_n$ converges. The series $\\sum_m \\abs{d_m} r^m$ is this series with zero terms inserted, so it converges too.\n4. By the [[term test|0VVY4H92G2]], $\\abs{d_m} r^m \\to 0$. So the sequence $(\\abs{d_m} r^m)$ is [[bounded|0S4JF619V7]].\n5. This holds for every $r \u003e 0$. By the [[definition of the radius of convergence|0660QX1VPR]], the radius of convergence of the cosine series is $\\infty$.\n6. By [[term-by-term differentiation|002X7EK91H]], $\\cos$ is differentiable on $\\mathbb{R}$, and $\\cos'(x) = \\sum_{m=1}^{\\infty} m d_m x^{m-1}$.\n7. The terms with odd $m$ vanish. For $m = 2n$ with $n \\ge 1$, the term is $2n \\frac{(-1)^n}{(2n)!} x^{2n-1} = \\frac{(-1)^n x^{2n-1}}{(2n-1)!}$.\n8. Put $k = n - 1 \\ge 0$. The term becomes $-\\frac{(-1)^k x^{2k+1}}{(2k+1)!}$.\n9. Removing the zero terms does not change the sum. So $\\cos'(x) = -\\sum_{k=0}^{\\infty} \\frac{(-1)^k x^{2k+1}}{(2k+1)!}$.\n10. By the definition of sine, this is $-\\sin x$. So $\\cos'(x) = -\\sin x$ for every real $x$.\n\n## Idea\n\nCosine is defined by a power series that converges everywhere, so it can be differentiated term by term. The constant term disappears, and each later term $\\frac{(-1)^n x^{2n}}{(2n)!}$ turns into a term of the sine series with the opposite sign.\n",
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