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How do we derive the addition formula of $\sin u$ from the following equation?

$$\frac{dx}{\sqrt{1 - x^2}} + \frac{dy}{\sqrt{1 - y^2}} = 0$$

Motivation

Let $u = \int_{0}^{x}\frac{dt}{\sqrt{1 - t^2}}$ Then $x = \sin u$

Let $v = \int_{0}^{y}\frac{dt}{\sqrt{1 - t^2}}$ Then $y = \sin v$

Let $u + v = const.$

Then $d(u + v) = \frac{dx}{\sqrt{1 - x^2}} + \frac{dy}{\sqrt{1 - y^2}} = 0$

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    This isn't exactly the same as what you're asking, but here's how to use differential equations to get the addition formula for sine. Fix a real number $y$. The function $x \mapsto \sin(x + y)$ satisfies the differential equation $f''(x) = -f(x)$ and so (by uniqueness of solutions to differential equations) is a linear combination $a\sin(x) + b\cos(x)$ for some real numbers $a$ and $b$ (which may depend on the constant $y$). We can solve for $a$ and $b$ by plugging in $x = 0$ (which gives us $b = \sin(y))$ and $x = -y$ (which gives us $a = \cos(y)$).2012-08-30
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    It’s probably my fault entirely, but I found your derivation of the differential equation and its relation to an addition formula unconvincing.2012-08-30
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    @Lubin I think the origin of the differential equation is probably Euler.2012-08-30
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    I noticed that someone serially upvoted for my questions and answers including this one. While I appreciate them, I would like to point out that serial upvotes are automatically reversed by the system.2013-11-27

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