Is this a valid proof of the Area of a Circle, assuming we do not already know $\pi$? In my AP Calculus class my friend and I decided we wanted to prove the area of a circle, without knowing $\pi$ to begin with. My friend and I each did it a different way. My method was to split the circle into a bunch of sectors, then find the area of each sector and multiply by the number of sectors to approximate the area of a circle. Here is what I came up with:
$f(r, n) = \frac{r^2 n \sin(\frac{180}{n})}{\sin(90 - \frac{180}{n})}$
With $r$ being the radius, and $n$ being the number of sides.
Then I rewrote it as
$f(r, n) =  \lim \limits_{n \to \infty} r^2  \frac{ n \sin(\frac{180}{n})}{\sin(90 - \frac{180}{n})}$
or
$f(r) =  r^2 \pi$
I have tested this many times and found it to work for finding the area of a circle. My question: Is my logic sound? Does this work without knowing $\pi$ beforehand?

The $180$ in your answer is really $\pi$ in disguise. The most natural definition of the sine function ends up using radian measure. Computing the sine in degrees then involves scaling by $\frac{\pi}{180}$. That is how your calculator is computing the sine (unless it is using lookup tables, which I highly doubt a modern calculator would be doing). In radians your limit becomes
$$\lim_{n\to\infty}{r^2\frac{n\sin{\frac{\pi}n}}{\sin{(\frac{\pi}2-\frac{\pi}{n})}}}$$
The denominator actually is not needed as it approaches $1$. So really the statement is that
$$\pi=\lim_{n\to\infty}{n\sin{\frac{\pi}{n}}}$$
This follows from the fact that
$$\lim_{x\to 0}{\frac{\sin{x}}{x}}=1$$
in radians.
In any case, if the question is whether your formula is correct, the answer is yes, and if there were some way of exactly calculating sine in degrees without knowing the value of $\pi$, then you wouldn't have to know the value of $\pi$. But I know of no such way.
