<s>[INST] Deriving the addition formula of $\sin u$ from a total differential equation 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 $d(u + v) = \frac{dx}{\sqrt{1 - x^2}} + \frac{dy}{\sqrt{1 - y^2}} = 0$
 [/INST] 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 $c$ be a constant.
$u + v = c$ is a solution of the equation:
$$\frac{dx}{\sqrt{1 - x^2}} + \frac{dy}{\sqrt{1 - y^2}} = 0$$
It suffices to prove that $\sin c = x\sqrt{1 - y^2} + y\sqrt{1 - x^2}$.
Since $v = c - u$, the right hand side is a function of $u$.
We write this function by $f(u)$.
Namely, $$f(u) = x\sqrt{1 - y^2} + y\sqrt{1 - x^2}$$
Let us calculate $\frac{df}{du}$.
$\frac{dx}{du} = 1/\frac{du}{dx} = \sqrt{1 - x^2}$
$\frac{dy}{du} = -\frac{dy}{dv} = -1/\frac{dv}{dy} = -\sqrt{1 - y^2}$
$\frac{d^2x}{du^2}= \frac{d\sqrt{1 - x^2}}{du}\cdot\frac{dx}{du} = \frac{-x}{\sqrt{1 - x^2}} \sqrt{1 - x^2} = -x$
$\frac{d^2y}{du^2}= \frac{d^2y}{dv^2} = -y$
Hence
$\frac{df}{du} = \frac{d}{du}(-x\frac{dy}{du} + y\frac{dx}{du})
= (-\frac{dx}{du}\frac{dy}{du} - x\frac{d^2y}{du^2}) + ( \frac{dy}{du}\frac{dx}{du}+y\frac{d^2x}{du^2}) = xy - yx = 0$
Hence $f(u)$ is constant.
Hence $f(u) = f(0) = y = \sin(v) = \sin(c)$ as desired. 
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