Finding sine of an angle in degrees without $\pi$ The following is found using a combination of: (a) a polygon with an infinite number of sides is a circle, (b) the perimeter of that polygon is the circumference of the circle that it becomes (of course), (c) the sine theorem, and (d) the ratio between the circumference of a circle and the diameter is $\pi$. 
$$\pi=\lim_{n\to\infty} n\sin \left(\frac{180}{n}\right)^o$$
I have been trying to rewrite the above expression without $sin$ in it explicitly. So, I researched other expressions for $sin$, and found this:
$$\sin (x) = x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\frac{x^9}{9!}-\frac{x^{11}}{11!}+\cdots$$
Attempting to rewrite this Taylor series in Sigma notation yielded this:
$$\sin(x)=\sum_{n=1}^{\infty} \frac{x^{2n-1}}{(2n-1)!}(-1)^{n-1}$$
The issue for me is that I have an angle measure in degrees, but $x$ above is in radians. I would convert the degrees to radians and be done with it just like that, except that I feel that solving for $\pi$ with $\pi$ is like defining a word with itself, and is 'cheating'.
I have been trying to find a way to make this conversion without $\pi$ but to no avail. I also tried using trig identities, but that has just led me in circles. 
Is there another way to find the sine of $x^o$ without involving $\pi$ at all?
An even more explicit and blunt form of my question: What is another way to express $sin(x^o)$ without using $\pi$?
To be clear, I am not looking for anything that involves approximations or infinite series that cannot be wholly expressed in a finite amount of space (ex. Taylor series above is okay because it can be expressed in sigma notation).
 A: As noted in a comment, there are equations you can write for the
trigonometric functions of rational portions of a right angle
(and therefore for the sine of any rational number of degrees)
without using $\pi$ or any trigonometric functions.
To actually use these equations may prove somewhat cumbersome, however.
Instead of attempting a formula for an arbitrary integer or rational
number of degrees, let me just address the desire to evaluate
$$
\newcommand{\Sin}{\mathop{\mathrm{Sin}}}
\newcommand{\Cos}{\mathop{\mathrm{Cos}}}
\lim_{n\to\infty} n \Sin\left(\frac{180}{n}\right)
$$
where $\Sin$ is the sine function that takes its parameter in degrees,
for example, $\Sin(90) = 1$.
It is sufficient to examine the following limit for integer values of $k$:
$$
\lim_{k\to\infty} 2^k \Sin\left(\frac{180}{2^k}\right).
$$
The quantity $\Sin\left(\frac{180}{2^k}\right)$ is easy to compute
(at least, it is easy compared to such values as $\Sin(1)$).
Let $\Cos(x)$ be the cosine function for $x$ measured in degrees;
then apply the half-angle formula for cosines of angles in the 
interval from $0$ to $180$ degrees, inclusive:
$$
\Cos\left(\frac{x}{2}\right) = \sqrt{\frac{1 + \Cos x}{2}}.
$$
If $x = \dfrac{180}{2^m}$, then $\dfrac x2 = \dfrac{180}{2^{m+1}}$
and the half-angle formula just says that
$$
\Cos\left(\frac{180}{2^{m+1}}\right)
 = \sqrt{\frac{1 + \Cos \left(\frac{180}{2^m}\right)}{2}}.
$$
You can find the value for any $k$ by starting at $m=1$ and applying the
half-angle formula repeatedly for $m=2,3,\ldots, k$:
\begin{align}
\Cos\left(\frac{180}{2}\right) &= \Cos(90) = 0,\\
\Cos\left(\frac{180}{4}\right) &=
 \sqrt{\frac{1 + \Cos \left(\frac{180}{2}\right)}{2}}
 = \sqrt{\frac{1 + 0}{2}} = \frac{1}{\sqrt 2}, \\
\Cos\left(\frac{180}{8}\right) &=
 \sqrt{\frac{1 + \Cos \left(\frac{180}{4}\right)}{2}}
= \sqrt{\frac{1 + \frac{1}{\sqrt 2}}{2}}
= \sqrt{\frac{\sqrt 2 + 1}{2\sqrt 2}}, \\
\Cos\left(\frac{180}{16}\right) &=
 \sqrt{\frac{1 + \Cos \left(\frac{180}{8}\right)}{2}}
= \sqrt{\frac{1 + \sqrt{\frac{\sqrt 2 + 1}{2\sqrt 2}}}{2}}
= \sqrt{\frac{\sqrt{2\sqrt 2} + \sqrt{\sqrt 2 + 1}}{2\sqrt{2\sqrt 2}}}, \\
\end{align}
and so forth to obtain $\Cos\left(\frac{180}{2^k}\right)$
for $k$ as large as desired.
Then obtain the sine by
$$
\Sin\left(\frac{180}{2^k}\right) =
\sqrt{1 - \left(\Cos\left(\frac{180}{2^k}\right)\right)^2}.
$$
A: In relation to David K's answer, I also note one small problem:
$$\sin(\frac x2)=\pm\sqrt{\frac{1-\cos(x)}2}$$
Obviously, $\sin(\frac x2)$ being negative is fully possible.
However, we know that for $0\le x\le\pi$, $\sin(x)>0$, in other words, it is positive.
On the interval from $-\frac{\pi}2\le x\le\frac{\pi}2$, $\cos(x)>0$.
So, we may start with $\cos(1)\approx0.540302306$.
We have $$\cos(0.5)\approx0.877581562$$
$$\cos(0.25)\approx0.968912422$$
$$\cos(0.125)\approx0.992197667$$
etc. (these values were obtained through the half angle formula)
Similarly, we have: (using half angle formula)
$$\sin(1)\approx0.841470985$$
$$\sin(0.5)\approx0.479425539$$
$$\sin(0.25)\approx0.247403959$$
$$\sin(0.125)\approx0.124674733$$
etc.
Now, knowing our trigonometric identities, one can use the sum of angles formula to approximate the desired value to desired accuracy.
