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  "body": "\n## Proof\n\nFirst we show that $2\\pi$ is a period. Then we locate the zeros of $\\sin$ in $(0, 2\\pi)$ and rule out smaller periods.\n\n1. From the [[defining series|0WB57D9488]], $\\sin 0 = 0$ and $\\cos 0 = 1$. By [[this result|01A8JGR4JM]], $\\cos \\pi = -1$ and $\\sin \\pi = 0$.\n2. By the [[addition formulas|028XEQCE88]] with $x = y = \\pi$, $\\cos 2\\pi = \\cos^2 \\pi - \\sin^2 \\pi = 1$ and $\\sin 2\\pi = 2 \\sin \\pi \\cos \\pi = 0$.\n3. By the addition formulas with $y = 2\\pi$, $\\sin(x + 2\\pi) = \\sin x \\cdot 1 + \\cos x \\cdot 0 = \\sin x$ and $\\cos(x + 2\\pi) = \\cos x \\cdot 1 - \\sin x \\cdot 0 = \\cos x$.\n4. By the [[definition of $\\pi$|0C1VYVNV2N]], $\\pi/2 \\in [0, 2]$ and $\\cos(\\pi/2) = 0$. Since $\\cos 0 = 1$, we get $0 \u003c \\pi/2 \\le 2$. So $2\\pi \u003e 0$ is a [[period|089KWVYD69]] of $\\sin$ and of $\\cos$.\n5. The sine series has only odd powers of $x$. Replacing $x$ by $-x$ changes the sign of every term, so $\\sin(-u) = -\\sin u$. Thus $\\sin$ is [[odd|0XE690QF65]].\n6. By the addition formulas and step 1, $\\sin(x + \\pi) = \\sin x \\cos \\pi + \\cos x \\sin \\pi = -\\sin x$ for all real $x$.\n7. Also $\\sin(\\pi - u) = \\sin \\pi \\cos(-u) + \\cos \\pi \\sin(-u) = -\\sin(-u) = \\sin u$ for all real $u$.\n8. The functions $\\sin$ and $\\cos$ are differentiable, by the [[derivative of sine|00R4MWW532]] and the [[derivative of cosine|0V2DV29EGT]]. So they are [[continuous|07ZMX4ANJS]].\n9. We claim $\\cos x \u003e 0$ for $0 \\le x \u003c \\pi/2$. Suppose $\\cos x \\le 0$ for such an $x$. Then $x \\ne 0$, and the [[intermediate value theorem|00KAFBR32E]] on $[0, x]$ gives $c \\in [0, x]$ with $\\cos c = 0$.\n10. Then $c \u003c \\pi/2 \\le 2$, so $c$ is a second zero of $\\cos$ in $[0, 2]$. This contradicts the definition of $\\pi$, and proves the claim.\n11. So $\\sin' = \\cos \u003e 0$ on $(0, \\pi/2)$. By [[this result|07J5X2T1ZX]], $\\sin$ is strictly increasing on $[0, \\pi/2]$. So $\\sin x \u003e \\sin 0 = 0$ for $0 \u003c x \\le \\pi/2$.\n12. By the [[Pythagorean identity|03H7WZ4T3K]], $\\sin^2(\\pi/2) = 1 - \\cos^2(\\pi/2) = 1$. By step 11, $\\sin(\\pi/2) = 1$.\n13. Let $\\pi/2 \u003c x \u003c \\pi$, and let $u = \\pi - x$. Then $0 \u003c u \u003c \\pi/2$, and by steps 7 and 11, $\\sin x = \\sin u \u003e 0$. So $\\sin \u003e 0$ on $(0, \\pi)$.\n14. Let $\\pi \u003c x \u003c 2\\pi$, and let $u = x - \\pi \\in (0, \\pi)$. By steps 6 and 13, $\\sin x = \\sin(u + \\pi) = -\\sin u \u003c 0$.\n15. So the only zero of $\\sin$ in the open interval $(0, 2\\pi)$ is $\\pi$.\n16. Suppose that $p$ is a period of $\\sin$ with $0 \u003c p \u003c 2\\pi$. Then $\\sin p = \\sin(0 + p) = \\sin 0 = 0$. By step 15, $p = \\pi$.\n17. But by steps 6 and 12, $\\sin(\\pi/2 + \\pi) = -\\sin(\\pi/2) = -1 \\ne 1 = \\sin(\\pi/2)$. So $\\pi$ is not a period of $\\sin$, a contradiction.\n18. Suppose that $p$ is a period of $\\cos$ with $0 \u003c p \u003c 2\\pi$. Then $\\cos(x + p) = \\cos x$ for all real $x$.\n19. Differentiating both sides with the [[chain rule|0RXX2PT1MQ]] gives $-\\sin(x + p) = -\\sin x$ for all real $x$. So $p$ is a period of $\\sin$, which contradicts steps 16 and 17.\n20. So $\\sin(x + 2\\pi) = \\sin x$ and $\\cos(x + 2\\pi) = \\cos x$ for all real $x$, and $2\\pi$ is the smallest positive period of $\\sin$ and of $\\cos$.\n\n## Idea\n\nThe values $\\cos 2\\pi = 1$ and $\\sin 2\\pi = 0$ make the addition formulas return the same values after a shift by $2\\pi$. To see that no shorter shift works, note that a period $p$ of sine must be a zero of sine. Inside $(0, 2\\pi)$ the only zero is $\\pi$, and shifting by $\\pi$ flips the sign of sine.\n",
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