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  "body": "\n## Proof\n\n1. Let $f(x) = \\sin^2 x + \\cos^2 x$ for real $x$, with $\\sin$ and $\\cos$ as [[defined by their power series|0WB57D9488]].\n2. By the [[derivative of sine|00R4MWW532]] and the [[derivative of cosine|0V2DV29EGT]], both functions are differentiable on $\\mathbb{R}$, with $\\sin' = \\cos$ and $\\cos' = -\\sin$.\n3. By the [[product rule|0V9A43H7WV]], $(\\sin^2)'(x) = 2 \\sin x \\cos x$ and $(\\cos^2)'(x) = -2 \\cos x \\sin x$.\n4. By [[linearity of the derivative|0M0TKFRX2A]], $f'(x) = 2 \\sin x \\cos x - 2 \\cos x \\sin x = 0$ for every real $x$.\n5. Since $\\mathbb{R}$ is an open interval, $f$ is [[constant|003QDZ4T16]].\n6. At $x = 0$, every term of the sine series vanishes, and only the first term of the cosine series is nonzero. So $\\sin 0 = 0$ and $\\cos 0 = 1$.\n7. So $f(0) = 0 + 1 = 1$, and $\\sin^2 x + \\cos^2 x = f(x) = 1$ for every real $x$.\n\n## Idea\n\nThe sum $\\sin^2 x + \\cos^2 x$ has derivative zero, because the two contributions $2 \\sin x \\cos x$ cancel. A function with zero derivative on $\\mathbb{R}$ is constant, and its value at $0$ is $1$.\n",
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