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

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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 $a = \cos(y)$). –  user29743 Aug 30 '12 at 22:48
It’s probably my fault entirely, but I found your derivation of the differential equation and its relation to an addition formula unconvincing. –  Lubin Aug 30 '12 at 23:29
@Lubin I think the origin of the differential equation is probably Euler. –  Makoto Kato Aug 30 '12 at 23:33
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. –  Makoto Kato Nov 27 '13 at 7:13

## 2 Answers

(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.

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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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I enjoyed this calculation. I would point out that to say $x = \sin(u)$ required the knowledge $\frac{d}{dx}\sin^{-1}(x) = \frac{1}{\sqrt{1-x^2}}$. In my usual presentation of calculus this derivation would be circular. That said, this will make a deliciously evil homework at some point in a future course I teach.... my students do not thank you. –  James S. Cook Sep 1 '12 at 2:12
@JamesS.Cook $u = \int_{0}^{x}\frac{dt}{\sqrt{1 - t^2}}$ is the arc length of the unit circle. So $x = \sin u$ is obvious, or $x = \sin u$ should be the definition of $\sin u$. –  Makoto Kato Sep 1 '12 at 2:23
with that definition of sine, I think that makes your argument non-circular. Very nice. I suppose the reason I would not use this definition is that I need a definition for sine which I can offer students of differential calculus. (I don't use Apostol, so derivatives come first) –  James S. Cook Sep 1 '12 at 13:48
@JamesS.Cook What is your definition of $\sin u$? –  Makoto Kato Sep 1 '12 at 17:12
Beginning with right-triangle trigonometry in quadrant I, I define $\sin (u)=y$. Then, I continue the sine function to the other quadrants by the same rule. Continuing for angles beyond $[0,2\pi]$ I insist on $2\pi$ periodicity of $\sin(u)$. I do think your definition is more elegant, but I need a definition that is accessible for students who have yet to study integral calculus. –  James S. Cook Sep 4 '12 at 1:43