TITLE: If we define $\sin x$ as series, how can we obtain the geometric meaning of $\sin x$?
QUESTION [29 upvotes]: In Terry Tao's textbook Analysis, he defines $\sin x$ as below:


Define rational numbers
Define Cauchy sequences of rational numbers, and equivalence of Cauchy sequences
Define reals as the space of Cauchy sequences of rationals modulo equivalence
Define limits (and other basic operations) in the reals
Cover a lot of foundational material including: complex numbers, power series, differentiation, and the complex exponential
Eventually (Chapter 15!) define the trigonometric functions via the complex exponential. Then show the equivalence to other definitions.


My question is how can we obtain the geometry interpretation of $\sin x$, that is, the ratio of opposite side and hypotenuse.

REPLY [3 votes]: The best approach I've seen uses the fact that $\exp z=\lim_{n\to\infty} (1+\frac{z}{n})^n$ for all complex $z$.  Then, for real $x$, we show that, for large $n$, $1+\frac{ix}{n}$, is "close enough" to $\cos \frac{x}{n} + i\sin\frac{x}{n}$, so that $(1+\frac{ix}{n})^n$ is "close enough" to $\cos x + i\sin x$.  (Here, $\cos$ and $\sin$ are the geometric definitions of the trig functions.)
So, first define $\exp z$ via the normal power series, and then show that $\exp z = \lim_{n\to\infty} (1+\frac{z}{n})^n$.
Then, define $\operatorname{cis} x =\cos x + i\sin x$, and prove, using (geometric) properties of the trig functions, that $\operatorname{cis} x\operatorname{cis} y =\operatorname{cis} (x+y)$. Then note that by induction, $(\operatorname{cis} x)^n = \operatorname{cis} nx$.
Next, we are going to compute $\exp(ix)$ by writing $1+\frac{ix}n$ in polar coordiates, $r_n \operatorname{cis}(\theta_n)$.
As mentioned above, the goal is to show that, for large enough $n$, $1+\frac{ix}{n}$ is "close enough" to $\operatorname{cis} \frac{x}n$, so that the limit of $(1+\frac{ix}n)^n$ is the same as the limit of $(\operatorname{cis}\frac x n)^n = \operatorname{cis} x$, and therefore $\exp(ix)=\operatorname{cis} x$.
First, note that $r_n=\sqrt{1+\frac{x^2}{n^2}}$, and show that $r_n^n\to 1$ as $n\to \infty$. (This is relatively easy, noting that $(r_n)^{2n^2}\to \exp (x^2)$.)
So this shows that $\exp(ix)$ ends up on the unit circle for real $x$, and is equal to $\lim_{n\to\infty} \operatorname{cis}(n\theta_n)$.
What we know about $\theta_n$ is that $\sin \theta_n = \frac{x}{nr_n}$.
Next you need to show is that $\lim_{n\to\infty} n\theta_n =  x$.  Then you've shown that $\exp(ix)=\operatorname{cis} x$.
The key to proving this last limit is to show that, for small $\theta$, $|\theta-\sin(\theta)|\leq C|\sin \theta|^2$ for some constant $C$.  Note, though, that proving this is not easy since we are assuming we do not know the power series expansion for $\sin$. 
If you can show that, then you can show, for large $n$, $|\theta_n - \frac{x}{nr_n}|\leq C|\frac{x}{r_nn}|^2<C|\frac{x}{n}|^2$. So $|n\theta_n - \frac{x}{r_n}|<C\frac{|x^2|}{n}$.
So, since $\frac{x}{r_n}\to x$, $n\theta_n\to x$, and therefore, $\lim \operatorname{cis} n\theta_n = \operatorname{cis} x$
So all that's left to prove is that for small enough $\theta$, $|\theta - \sin \theta| \leq C|\sin \theta|^2$. (You could actually replace $2$ with $1+\epsilon$ for fixed $\epsilon>0$.)
The geometric reason for this last one is as follows. Let $P=(1,0)$ and $Q=(\cos \theta,\sin \theta)$.  Take the lines tangent to the circle at $P$ and $Q$ and let their intersection be $R$. Claims (for small $\theta>0$:)
$$|PR|=|QR|=\tan\frac{\theta}2$$
$$\sin\theta < \theta < |PR|+|QR|=2\tan\frac{\theta}2$$
The first half of the inequality is because $\sin \theta$ is the shortest distant from $Q$ to the $x$-axis, and $\theta$ is the length of the path from $Q$ to the $x$-axis along the circle.  The second inequality is a little less intuitive - basically, this is due to the rule that the shortest path from $P$ to $Q$ that does not go inside the circle is along the circle.  
(For small $\theta<0$, we have to reverse all the signs, but the results are the same.)
Thus we see:
$$|\theta-\sin \theta| < |2\tan\frac\theta 2 - \sin\theta|$$
But $2\tan\frac{\theta}2 = \frac{2\sin \frac{\theta}2}{\cos\frac\theta 2}$.  Mutliply numerator and denominator by $\cos \frac{\theta}2$ and we see that:
$$2\tan\frac{\theta}2 = \frac{\sin \theta}{\cos^2\frac{\theta}2}$$
So:
$$|2\tan\frac\theta 2 - \sin\theta| = |\sin \theta \frac{\sin^2\frac \theta 2}{\cos^2\frac\theta 2}|$$
Multiplying the numerator and denominator by $4\cos^2\frac \theta 2$, we get:
$$|2\tan\frac\theta 2 - \sin\theta| = |\sin\theta \frac{\sin^2\theta}{4\cos^4\frac{\theta}2}|$$
So with $\theta$ small enough that $\cos\frac\theta 2>\frac{1}{2}$, we have 
$$|\theta-\sin \theta| < |2\tan\frac\theta 2 - \sin\theta| < 4|\sin^3\theta|<4|\theta|^3$$
So we now know that $\exp ix = \operatorname{cis} x$.  That means, in turn, that $$\sin x = \frac{\exp(ix)-\exp(-ix)}{2i}$$ and $$\cos x = \frac{\exp(ix)+\exp(-ix)}{2}$$
From there we can derive the power series for $\sin$ and $\cos$ using the power series for $\exp$.