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:

1. Define rational numbers
2. Define Cauchy sequences of rational numbers, and equivalence of Cauchy sequences
3. Define reals as the space of Cauchy sequences of rationals modulo equivalence
4. Define limits (and other basic operations) in the reals
5. Cover a lot of foundational material including: complex numbers, power series, differentiation, and the complex exponential
6. 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.

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.

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.

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.

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.