In this approximation of $\pi$, do you need to know $\pi$ make these calculations? $\pi = 2n\dfrac{\cos (x)}{\sin (x)+1}$
where $x = 90°\dfrac{n-2}{n}$
and $n \to \infty$
A high school student came up with the idea for this approximation of $\pi$, and I helped develop it. It is based on an inscribed polygon. Is this a circular definition? Does it require knowledge of the value of $\pi$ to work?
 A: With some trigonometric identities we can rewrite this as $\pi=\lim_{n\to\infty}n\tan\frac{\pi}{n}$, which doesn't require knowledge of $\pi$ provided we consider values of $n$ that are powers of $2$. The insight is that $\tan 2x =\frac{2\tan x}{1-\tan^2 x}$ implies $$\tan 2x\tan^2 x +2\tan x - \tan 2x=0,\,\tan x =\frac{-1+\sqrt{1+\tan^2 2x}}{\tan 2x}$$for small $x>0$. (The sign used in the quadratic formula follows from $\tan 2x \approx 2x,\,\tan x \approx x.$) Now use $\tan\frac{\pi}{4}=1$ to compute $\tan\frac{\pi}{2^k}$ for $k\ge 3$. Whereas the case $k=2$ gives $\pi\approx 4\cdot 1 = 4$, $k=3$ gives $\pi\approx 8\cdot(\sqrt{2}-1)\approx 3.3$.
How good is this approximation? Writing $n=2^k$ we have $$n\tan\frac{\pi}{n}\approx n(\frac{\pi}{n}+\frac{1}{3}(\frac{\pi}{n})^3)=\pi+\frac{\pi^3}{3n^2}=\pi+\frac{\pi^3}{3\cdot 4^k},$$so to get $d$ decimal places right requires $k\approx d\dfrac{\log 10}{\log 4}$.
A: Seriously edited:
As $x$ in your formula is in degrees, it depends on whether you can calculate $\sin$ and $\cos$ to something depending on $x$ without knowing $\pi$. In general I don't think that's possible, but as crivair points out in a comment both below the question and below this answer, we can prove that the expression can be calculated (nothing said about how easy though) for certain $n$'s.
And then a word about terminology:
$\pi$ has a well known and short definition: The ratio between the circumference and diameter of a circle!
You should not go about inventing other definitions, that's like defining that your apple is blue, so this should not be considered a definition, but it is a limit that could theoretically (but as pointed out, in practice it might not be very good) to calculate $\pi$.
With the terminology in place: The only methods I know for calculating $\sin$ and $\cos$ uses infinite series (and assume that argument is in radians, and you need to know $\pi$ to convert between degrees and radians), if that's all you can find that will work in your case, you'll have two infinite sums, so getting a good value for $\pi$ from your limit won't just require choosing $n$ large enough, it also requires you to calculate those sums to a sufficiently high precision.
