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  "body": "\n## Proof\n\n1. By the [[definition of sine|0WB57D9488]], $\\sin x = \\sum_{m=0}^{\\infty} c_m x^m$, where $c_{2n+1} = \\frac{(-1)^n}{(2n+1)!}$ and $c_{2n} = 0$ for $n \\ge 0$. Here $k!$ is the [[factorial|0WFCGVMRRC]].\n2. Let $r \u003e 0$, and let $t_n = \\frac{r^{2n+1}}{(2n+1)!} \u003e 0$. Then $\\frac{t_{n+1}}{t_n} = \\frac{r^2}{(2n+2)(2n+3)} \\to 0$.\n3. By the [[ratio test|0QNY6ZBXPN]], $\\sum_n t_n$ converges. The series $\\sum_m \\abs{c_m} r^m$ is this series with zero terms inserted, so it converges too.\n4. By the [[term test|0VVY4H92G2]], $\\abs{c_m} r^m \\to 0$. So the sequence $(\\abs{c_m} r^m)$ 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 sine series is $\\infty$.\n6. By [[term-by-term differentiation|002X7EK91H]], $\\sin$ is differentiable on $\\mathbb{R}$, and $\\sin'(x) = \\sum_{m=1}^{\\infty} m c_m x^{m-1}$.\n7. The terms with even $m$ vanish. For $m = 2n + 1$ the term is $(2n+1) \\frac{(-1)^n}{(2n+1)!} x^{2n} = \\frac{(-1)^n x^{2n}}{(2n)!}$.\n8. Removing the zero terms does not change the sum. So $\\sin'(x) = \\sum_{n=0}^{\\infty} \\frac{(-1)^n x^{2n}}{(2n)!}$.\n9. By the definition of cosine, this sum is $\\cos x$. So $\\sin'(x) = \\cos x$ for every real $x$.\n\n## Idea\n\nSine is defined by a power series that converges everywhere, and such a series can be differentiated term by term. Differentiating $\\frac{x^{2n+1}}{(2n+1)!}$ gives $\\frac{x^{2n}}{(2n)!}$, so each term of the sine series becomes the matching term of the cosine series.\n",
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