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).
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 $$ \Sin\left(\frac{180}{2^k}\right) = \sqrt{1 - \left(\Cos\left(\frac{180}{2^k}\right)\right)^2}. $$