If we define $\sin x$ as series, how can we obtain the geometric meaning of $\sin x$? 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.
 A: To get to the geometry, take the "non-geometric" versions of cosine and sine, say $C(t)$ and $S(t)$. We can use these to parametrize the unit circle, simply because of the fact that $C^2(t)+S^2(t)=1$.  
Now comes the crucial bit, calculating the arclength. We find that the arclength from $0$ to $t$ is $t$.  This connects $C(t)$ and $S(t)$ with "angle," which in the formal theory is just arclength. 
A: 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 $\exp$.
A: Knowing that $(\cos x)'=-\sin x$ and that $(\sin x)'=\cos x$ (which I assume Tao proves) allows one to show that for $f(x)=\sin^2 x+\cos^2 x$, we have 
$$f'(x)=2\sin x \cos x-2\cos x\sin x =0.$$ Thus, $f$ is a constant function. Since $f(0)=1$, $f$ is identically 1. 
So, the Pythagorean identity is valid:
$$
\sin^2 x+\cos^2x=1.
$$
Using this, it follows that the curve $C$ parameterized by $x=\cos t$, $y=\sin t$ is a circle of radius 1 centered at the origin. Further analysis of this curve will reveal that the values of $\sin$ and $\cos$ can be read from the side lengths of right triangles. 
Note, in particular that the length of arc from $t=0$ to $t=t_0$ is 
$\int_0^{t_0} \sqrt{\bigl[{dx\over dt} \bigr]^2 +\bigl[{dy\over dt}\bigr]^2 }\,dt=t_0$; so the  angle to which the "triangle method" refers is interpreted in the correct manner.
A: In this hint I suggest showing from the power series that if
$$
\sin(x)=\sum_{k=0}(-1)^k\frac{x^{2k+1}}{(2k+1)!}\tag{1}
$$
and
$$
\cos(x)=\frac{\mathrm{d}}{\mathrm{d}x}\sin(x)=\sum_{k=0}(-1)^k\frac{x^{2k}}{(2k)!}\tag{2}
$$
that $\frac{\mathrm{d}}{\mathrm{d}x}\cos(x)=-\sin(x)$ and from there that 
$$
\sin^2(x)+\cos^2(x)=1\tag{3}
$$
Therefore, $(\cos(x),\sin(x))$ lies on the unit circle.
To see that $(\cos(x),\sin(x))$ moves around the unit circle at unit speed, note that $(3)$ implies
$$
\left|\frac{\mathrm{d}}{\mathrm{d}x}(\cos(x),\sin(x))\right|=\left|(-\sin(x),\cos(x))\right|=1\tag{4}
$$
Thus, $(3)$ and $(4)$ say that $(\cos(x),\sin(x))$ moves around the unit circle at unit speed. Note also that $(-\sin(x),\cos(x))$ is at a right angle counter-clockwise from $(\cos(x),\sin(x))$. Therefore, $(\cos(x),\sin(x))$ moves counter-clockwise around the unit circle at unit speed, starting at $(1,0)$. This should be sufficient to show that $\sin(x)$ and $\cos(x)$ are the standard trigonometric functions.
A: From the series, it is easy to see Euler's forumla,
$$ e^{ix} = \cos(x) + i\sin(x)$$
With more series manipulation, we can obtain the Pythagorean theorem,
$$|e^{ix}| = e^{ix}e^{-ix} = (\cos(x) + i\sin(x))(\cos(x) - i\sin(x)) = \cos^{2}(x) + \sin^{2}(x) = 1$$
Knowing that $\sin(x)$ and $\cos(x)$ have range $[-1,1]$, and are odd and even functions respectively, we see that $e^{ix}$ traces out the unit circle in $\mathbb{C}$. From this, we can extract the geometric interpretation of sine and cosine. 
A: Euler's formula (the complex exponential) and the identification of the complex plane $\mathbb{C}$ with $\mathbb{R}^2$ does that for us. Then the equations $x=\cos\theta,~y=\sin\theta$ define the ordinate and abscissa of points on the unit circle; drawing in the relevant triangles gets you the rest of the way.
From the series definition of $\sin$ and $\cos$, the easiest link with geometry is via the complex exponential series to derive Euler's formula, $\cos\theta+i\sin\theta=e^{i\theta}$. Once we have that, the rest follows (from the background that Tao provides).
