Pendulum (highschool) I have a question concerning pendulum.
On our physics lessons we built simple pendulum consisting of weight and a thread ($\ell=0.31$ m).
We tried to calculate $g$ from the well known formula 
$$T_m = 2\pi\sqrt{\frac{\ell}{g}}$$
which leads to 
$$g = \frac{4\pi^2 \ell}{T_m^2}$$
Of course, this formula comes from conclusion that $\sin(\alpha)-\tan(\alpha)$ is neglectable for "small" $\alpha$.
Hence our target was to find formula, which works for larger $\alpha$ as well.
According to this two sources, this should do the trick:
$$T_m = 2\pi\sqrt{\frac{\ell}{g}}(1+\frac{1}{4}\sin^2(\frac{\alpha}{2}))$$
from which I came to this conclusion:
$$g=\frac{\ell\pi^2(\cos(\alpha)-9)^2}{16T_m^2}$$
Here is my problem: After using this formula and our measured data, I got $\overline g = 8.9085$, which is obviously not very accurate (for ~48°N).
Any ideas where I've made a mistake?
 A: UPDATE
I finally got it right. My thanks goes to @NeuroFuzzy, who pointed me in the right direction.
According to wiki's Legendre polynomial solution for the elliptic integral, "an exact solution to the period of a pendulum is:"
$$T=2\pi\sqrt\frac{\ell}{g}\sum\limits_{n=0}^\infty \left[\left(\frac{(2n)!}{(2^nn!)^2}\right)^2sin^{2n}\left(\frac{\alpha}{2}\right)\right]$$
which after solving for g gives
$$g=\ell\left(2\pi \sum\limits_{n=0}^\infty \left[\left(\frac{(2n)!}{(2^nn!)^2}\right)^2sin^{2n}\left(\frac{\alpha}{2}\right)\right]T^{-1} \right)^2$$
Although it is probably not the most beautiful wonder of the world, it seems to work for me. I finally get acceptable results ($\overline g = 9.402$). The difference between real and this g can easily be explained, therefore I mark this post as the real answer. In case somebody is interested in this, here is python2 code for finding g from measured data.
Original answer
After reading your answers, suggestions and after going through simplification of the formula once again (with the very same result), I have to conclude that, assuming
$$T_m = 2\pi\sqrt{\frac{\ell}{g}}(1+\frac{1}{4}\sin^2(\frac{\alpha}{2}))$$
is correct (taken from two independent sources), following formula is correct as well
$$g=\frac{\ell\pi^2(\cos(\alpha)-9)^2}{16T_m^2}$$
This suggests my experiment alone was not very accurate, thus resulting in the error seen in my original question.
Additional info
I have tried to come up with the formula for $T_m$ but with no luck. I got
$$T=2\pi\sqrt{\frac{\ell\cos(\alpha)}{g}}$$
from which one can easily get
$$g=\frac{4\ell\pi^2\cos(\alpha)}{T^2}$$
Although it looks little bit similar, it's not quite there. It also returns unacceptable results ($\overline g = 4.94$)
A: This is really a comment to Mathbreaker's answer, but it's hard to do formulae in comments.
If you simply solve:
$$ T_m = 2\pi\sqrt{\frac{\ell}{g}}(1+\frac{1}{4}\sin^2(\tfrac{\alpha}{2})) $$
for $g$ you get:
$$ g = \frac{4 \pi^2 l (4 + \sin^2(\tfrac{\alpha}{2}))^2}{16 \tau^2} \tag{1} $$
We use the identity:
$$ \sin^2(\tfrac{\alpha}{2}) = \tfrac{1}{2} - \tfrac{1}{2}\cos\alpha $$
Substituting this in (1) gives:
$$\begin{align}
 g &= \frac{4 \pi^2 l (4 + \tfrac{1}{2} - \tfrac{1}{2}\cos\alpha)^2}{16 \tau^2} \\
   &= \frac{\pi^2 l (9 - \cos\alpha)^2}{16 \tau^2}
\end{align}$$
which is kecer's formula (give or take a factor of -1 that gets squared anyway).
A: $$g= \frac{L *4\pi^2 * ((5- \cos^2 (\alpha/2))^2}{16T^2}$$ 
This is what i got when i simplified for $g$.
I assume you made some mistake in simplification, or it just might be me, Iapologize if it is. You could try plugging your values into this and hope for an answer... 
when i simplify the latter part as well i get 
$$\frac{L * 4\pi^2 *(25+\cos^4(\alpha/2) - 10\cos^2(\alpha/2))}{16T^2}$$
Either there is a mistake in the formula you have obtained online, or in simplification. 
Also, there is a possibility that your measure data wouldve been inaccurate due to parallax error or otherwise.
