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  "body": "\n## Proof\n\n1. From the [[defining series|0WB57D9488]], $\\sin 0 = 0$ and $\\cos 0 = 1$. The sine series has only odd powers and the cosine series only even powers.\n2. So replacing $x$ by $-x$ gives $\\sin(-x) = -\\sin x$ and $\\cos(-x) = \\cos x$. That is, $\\sin$ is [[odd and $\\cos$ is even|0XE690QF65]].\n3. By the [[derivative of sine|00R4MWW532]] and the [[derivative of cosine|0V2DV29EGT]], both functions are differentiable on $\\mathbb{R}$, hence [[continuous|07ZMX4ANJS]].\n4. By the [[definition of $\\pi$|0C1VYVNV2N]], $\\pi/2$ is the only zero of $\\cos$ in $[0, 2]$. Since $\\cos 0 = 1$, we get $0 \u003c \\pi/2 \\le 2$.\n5. 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 $\\cos c = 0$.\n6. Then $c \u003c \\pi/2 \\le 2$, so $c$ is a second zero of $\\cos$ in $[0, 2]$, a contradiction. This proves the claim.\n7. Since $\\cos$ is even, $\\cos y \u003e 0$ for every $y$ in $I = (-\\pi/2, \\pi/2)$.\n8. On $[-\\pi/2, \\pi/2]$, $\\sin$ is continuous with $\\sin' = \\cos \u003e 0$ at interior points. By [[this result|07J5X2T1ZX]], $\\sin$ is strictly increasing on $[-\\pi/2, \\pi/2]$.\n9. So $\\sin(\\pi/2) \u003e \\sin 0 = 0$. By the [[Pythagorean identity|03H7WZ4T3K]], $\\sin^2(\\pi/2) = 1 - \\cos^2(\\pi/2) = 1$. So $\\sin(\\pi/2) = 1$, and $\\sin(-\\pi/2) = -1$ by oddness.\n10. By steps 8 and 9, $\\sin$ maps $I$ into $(-1, 1)$. Let $f \\colon I \\to \\mathbb{R}$ be the restriction of $\\sin$ to the open interval $I$.\n11. Let $-1 \u003c x \u003c 1$, and let $y = \\arcsin x$. By the [[definition of arcsine|0B89D93YA1]], $y \\in [-\\pi/2, \\pi/2]$ and $\\sin y = x$.\n12. Since $\\sin(\\pm \\pi/2) = \\pm 1 \\ne x$, we have $y \\in I$. So $f(I) = (-1, 1)$, and the inverse $f^{-1} \\colon (-1, 1) \\to I$ equals $\\arcsin$ on $(-1, 1)$.\n13. The function $f$ is continuous and strictly monotone on $I$. It is differentiable at $y$ with $f'(y) = \\cos y \u003e 0$ by step 7.\n14. By the [[inverse function rule|0M0RZ3HSB0]], $\\arcsin$ is differentiable at $x = f(y)$, and $\\arcsin'(x) = \\frac{1}{\\cos y}$.\n15. By the Pythagorean identity, $\\cos^2 y = 1 - \\sin^2 y = 1 - x^2$. Since $\\cos y \u003e 0$, we get $\\cos y = \\sqrt{1 - x^2}$.\n16. So $\\arcsin$ is differentiable on $(-1, 1)$, and $\\frac{d}{dx} \\arcsin x = \\frac{1}{\\sqrt{1 - x^2}}$ for $-1 \u003c x \u003c 1$.\n\n## Idea\n\nOn $(-\\pi/2, \\pi/2)$ sine is strictly increasing with derivative $\\cos y \u003e 0$, so its inverse has derivative $\\frac{1}{\\cos y}$. Since $\\sin y = x$ and $\\cos y \u003e 0$ there, the Pythagorean identity rewrites $\\cos y$ as $\\sqrt{1 - x^2}$.\n",
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