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  "body": "\n## Proof\n\n1. From the [[defining series|0WB57D9488]], $\\sin 0 = 0$ and $\\cos 0 = 1$. At $0$ every sine term vanishes, and only the first cosine term is nonzero.\n2. By the [[definition of $\\pi$|0C1VYVNV2N]], $\\pi/2$ is the only zero of $\\cos$ in $[0, 2]$. Since $\\cos 0 = 1 \\ne 0$, we get $0 \u003c \\pi/2 \\le 2$.\n3. By the [[derivatives of cosine|0V2DV29EGT]] and [[of sine|00R4MWW532]], both functions are differentiable on $\\mathbb{R}$. So they are [[continuous|07ZMX4ANJS]].\n4. We claim $\\cos x \u003e 0$ for $0 \\le x \u003c \\pi/2$. Suppose instead that $\\cos x \\le 0$ for such an $x$. Then $x \\ne 0$, since $\\cos 0 = 1$.\n5. The value $0$ lies between $\\cos 0 = 1$ and $\\cos x$. By the [[intermediate value theorem|00KAFBR32E]] on $[0, x]$, there is $c \\in [0, x]$ with $\\cos c = 0$.\n6. Then $c \\le x \u003c \\pi/2 \\le 2$. So $c$ is a zero of $\\cos$ in $[0, 2]$ other than $\\pi/2$, a contradiction. This proves the claim.\n7. On $[0, \\pi/2]$, $\\sin$ is continuous, and $\\sin' = \\cos \u003e 0$ at each interior point. By [[this result|07J5X2T1ZX]], $\\sin$ is strictly increasing on $[0, \\pi/2]$.\n8. So $\\sin(\\pi/2) \u003e \\sin 0 = 0$.\n9. By the [[Pythagorean identity|03H7WZ4T3K]], $\\sin^2(\\pi/2) = 1 - \\cos^2(\\pi/2) = 1$. With step 8, $\\sin(\\pi/2) = 1$.\n10. By the [[addition formulas|028XEQCE88]] with $x = y = \\pi/2$, $\\cos \\pi = \\cos^2(\\pi/2) - \\sin^2(\\pi/2) = 0 - 1 = -1$.\n11. In the same way, $\\sin \\pi = 2 \\sin(\\pi/2) \\cos(\\pi/2) = 2 \\cdot 1 \\cdot 0 = 0$.\n\n## Idea\n\nCosine is positive before its first zero $\\pi/2$, so sine increases there and $\\sin(\\pi/2)$ is positive. The Pythagorean identity then forces $\\sin(\\pi/2) = 1$. Doubling the angle with the addition formulas gives the values at $\\pi$.\n",
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