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  "body": "\n## Proof\n\nWe fix $y$ and show that the differences of the two sides vanish, using derivatives.\n\n1. Fix a real number $y$. For real $x$ let $g(x) = \\sin(x + y) - \\sin x \\cos y - \\cos x \\sin y$.\n2. For real $x$ let $h(x) = \\cos(x + y) - \\cos x \\cos y + \\sin x \\sin y$.\n3. The map $x \\mapsto x + y$ has derivative $1$. By the [[chain rule|0RXX2PT1MQ]] with the [[derivative of sine|00R4MWW532]], $\\frac{d}{dx} \\sin(x + y) = \\cos(x + y)$.\n4. In the same way, the [[derivative of cosine|0V2DV29EGT]] gives $\\frac{d}{dx} \\cos(x + y) = -\\sin(x + y)$.\n5. By [[linearity of the derivative|0M0TKFRX2A]], with $\\cos y$ and $\\sin y$ constants, $g'(x) = \\cos(x + y) - \\cos x \\cos y + \\sin x \\sin y = h(x)$.\n6. In the same way, $h'(x) = -\\sin(x + y) + \\sin x \\cos y + \\cos x \\sin y = -g(x)$.\n7. Let $q(x) = g(x)^2 + h(x)^2$. By the [[product rule|0V9A43H7WV]], $q'(x) = 2 g(x) h(x) + 2 h(x) (-g(x)) = 0$ for every real $x$.\n8. Since $\\mathbb{R}$ is an open interval, $q$ is [[constant|003QDZ4T16]].\n9. From the [[defining series|0WB57D9488]], $\\sin 0 = 0$ and $\\cos 0 = 1$. So $g(0) = \\sin y - 0 - \\sin y = 0$ and $h(0) = \\cos y - \\cos y + 0 = 0$.\n10. So $q(x) = q(0) = 0$ for every $x$. A sum of two squares of reals is $0$ only if both are $0$. So $g(x) = 0$ and $h(x) = 0$.\n11. Since $y$ was arbitrary, $\\sin(x + y) = \\sin x \\cos y + \\cos x \\sin y$ and $\\cos(x + y) = \\cos x \\cos y - \\sin x \\sin y$ for all real $x$ and $y$.\n\n## Idea\n\nThe two error terms $g$ and $h$ satisfy $g' = h$ and $h' = -g$, like a rotation. So the quantity $g^2 + h^2$ cannot change. It starts at $0$ when $x = 0$, so both errors stay $0$ forever.\n",
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