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 $d(u + v) = \frac{dx}{\sqrt{1 - x^2}} + \frac{dy}{\sqrt{1 - y^2}} = 0$
 A: (this is not really an answer, but I spent a while thinking about your question and this comment is too long for a comment!)
Let me explain a few concerns I have concerning the possible circularity of the proposed derivation. If the derivation involves:
$$ \frac{d}{dx} (\sin(x)) = \cos(x) $$
then this begs the question: how is this known? If we adopt the standard approach,
$$ \begin{align}
\frac{d}{dx} (\sin(x)) &= \lim_{h \rightarrow 0} \frac{ \sin(x+h)-\sin(x)}{h} \\
&=\lim_{h \rightarrow 0} \frac{ \sin(x)\cos(h)+\sin(h)\cos(x)-\sin(x)}{h} \\
&=\sin(x)\lim_{h \rightarrow 0} \frac{ \cos(h)-1}{h}+\cos(x)\lim_{h \rightarrow 0}\frac{\sin(h)}{h} \\
&= \sin(x)(0)+\cos(x)(1) \\
&=\cos(x).
\end{align} $$
then we implicitly use of the sine angle addition formula every time we take a derivative of sine. In order to make the proposed derivation non-circular we need a way to show $\frac{d}{dx} \sin(x) = \cos(x)$ without using the adding-angles formula for sine.
A connected comment; applying L'Hopital's Rule to obtain $\lim_{h \rightarrow 0} \frac{\sin(h)}{h} = \lim_{h \rightarrow 0} \frac{\cos(h)}{1} = \cos(0)=1$ is circular. I need the limit $\lim_{h \rightarrow 0} \frac{\sin(h)}{h}=1$ in order to show the derivative of sine is cosine. 
Obviously, if you can supply a definition of sine and cosine which is divorced from the adding angles formula then the proposed derivation becomes more interesting. 
I'll follow countinghaus's comment and twist it a bit. One alternative way to define sine and cosine is as the solutions of the ODE $y''+y=0$. If we say sine is the odd solution and cosine is the even solution then you can derive the standard identities for sine and cosine via their presentation as power series(which are easily derived from $y''+y=0$). I read this in Edward's Advanced Calculus text, I bet it can be found elsewhere. Continuing in this thought, once we label the power series $\sum_{n=0}^{\infty}\frac{(-1)^nx^{2n}}{(2n)!}$ as $\cos(x)$ and $\sum_{n=0}^{\infty}\frac{(-1)^nx^{2n+1}}{(2n+1)!}$ as $\sin(x)$ then we can follow countinghaus's comment and derive the adding angles formula through his argument.
The argument outined above is intriguing because it does not appear to use analytic geometry. In contrast, the usual approach I take is to define $\sin(x),\cos(x)$ by extending right-triangle trigonometry to make polar coordinates formulas work. Then, the adding angles formulas are derived through arguments of analytic geometry. In particular, if you apply the law of cosines to a triangle with angle $a-b$ inscribed in the unit-circle then the formulas $\cos(a-b)=\cos(a)\cos(b)+\sin(a)\sin(b)$. Once that is known, the usual addition of angle formulas fall out easily.
In summary: I really want to see this derivation you propose, but, perhaps it should be done with the understanding that sine and cosine are defined in some way divorced from direct analytic geometry. 
A: 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. 
