$\pi = 2n\dfrac{\cos (x)}{\sin (x)+1}$
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?
On
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 $k\approx d\dfrac{\log 10}{\log 4}$.
On
this looks kinda like the Archimedes $\pi$ calculation! First off, since
$$\begin{align}\sin\left(90°\frac{n-2}n\right)&=\sin\left(90°-90°\frac2n\right)=\cos\left(90°\frac2n\right)=\cos\left(\frac{\pi}n\right)\\
\cos\left(90°\frac{n-2}n\right)&=\cos\left(90°-90°\frac2n\right)=\sin\left(90°\frac2n\right)=\sin\left(\frac{\pi}n\right)\end{align}$$
So I am going to write your formula as
$$\pi=2n\frac{\sin\left(\frac{\pi}n\right)}{1+\cos\left(\frac{\pi}n\right)}=2n\tan\left(\frac{\pi}{2n}\right)$$
Let'a look at a picture of what this means geometrically:
In the figure above, $A+B$ is the area of sector $AQT$ of the unit circle of angle $\theta$, so $A+B=\frac12\theta$. $A$ is the area of triangle $OPT$ so it has area $\frac12\sin\theta\cos\theta$. $A+B+C+D+E$ is the area of triangle $OQS$ so it has area $\frac12\tan\theta$. By containment $$\frac12\sin\theta\cos\theta\le\frac12\theta\le\frac12\tan\theta$$ If we multiply through by $4$, we have $$2\sin\theta\cos\theta=\sin2\theta\le2\theta\le2\tan\theta$$ Replacing $\theta$ by $\frac{\theta}2$, we get to the result we have been aiming at: $$\sin\theta\le\theta\le2\tan\left(\frac{\theta}2\right)$$ So what your formula does it to approximate $\frac{\pi}n\approx2\tan\left(\frac{\theta}2\right)$. So it's a reasonable approximation and the order of magnitude of the original Archimedes one, but just looking at the figure we can do better. Consider the parabola $$x=1-\frac12\sec^2\left(\frac{\theta}2\right)y^2$$ Where $x$ is the horizontal distance from $O$ and $y$ is the vertical distance from $O$ hence also $Q$. You can barely see the sliver of area $C$ between the parabolic arc $QT$ and circular arc $QT$ and area triangle $TRS$ has area $E$ of $$\frac12(1-\cos\theta)(\tan\theta-\sin\theta)=\frac12\tan\theta(1-\cos\theta)^2=2\tan\theta\sin^4\left(\frac{\theta}2\right)$$ So neglecting areas $C$ and $E$ we have from The mensuration formula for a parabolic arc $$B\approx B+C=2D\approx2(C+D+E)$$ $$\begin{align}\frac14\sin2\theta+2D&=A+B+C\approx A+B=\frac{\theta}2\\ &=A+B+C+D+E-(C+D+E)\\ &\approx A+B+C+D+E-D\\ &=\frac12\tan\theta-D\end{align}$$ So then $$\frac14\sin2\theta\left(\frac12\tan\theta\right)^2\approx\left(\frac{\theta}2-2D\right)\left(\frac{\theta}2+D\right)^2=\frac18\theta^3-\frac32D^2\theta-2D^3$$ So on replacing $\theta$ by $\frac{\theta}2$ once again we see that $$\theta\approx\sqrt[3]{4\sin\theta\tan^2\left(\frac{\theta}2\right)}$$ Should be a better approximation. Let's see how this works in practice: $$\begin{array}{r|l|ll}n&\theta&2n\tan\frac{\theta}2&n\sqrt[3]{4\sin\theta\tan^2\frac{\theta}2}\\\hline2&1.570796&4.00000000&3.174802103936399\\ 4&0.785398&3.31370850&3.143338840697821\\ 8&0.392699&3.18259788&3.141697707414082\\ 16&0.196350&3.15172491&3.141599158903353\\ 32&0.098175&3.14411839&3.141593059237774\\ 64&0.049087&3.14222363&3.141592678928243\\ 128&0.024544&3.14175037&3.141592655173220\\ 256&0.012272&3.14163208&3.141592653688754\\ 512&0.006136&3.14160251&3.141592653595979\\ 1024&0.003068&3.14159512&3.141592653590180\\ \end{array}$$ So this consideration of the relative sizes of areas $B$ and $C+D+E$ earned us about twice as many significant figures.
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.