How can I get a "better small angle approximation"? Is there a way to improve on the small angle approximation so that I get a more accurate answer. Is there any way to use a higher order series for the $\sin$ function or to use some sort of series approximation to the elliptical solution to the equation?
For reference, the equation is $$y'' = -\sin(y).$$
 A: If you use the third order Taylor approximation and solve the equation with constants $c_1$ and $c_2$ you can find a solution with the Elliptic F function rather than amplitude. For convience, let $c_3 = \sqrt{9-3c_1}$.
$$-\frac{6 \left(2 c_3-y^2+6\right) \left(c_3+y^2-6\right) F\left(i \sinh ^{-1}\left(y\sqrt{\frac{1}{2 c_3-6}} \right)|\frac{3-c_3}{c_3+3}\right){}^2}{\left(c_3+3\right) \left(12 c_1+y^4-12 y^2\right)}=\left(c_2+x\right){}^2$$
With a fifth order approximation the solution is so bad that I am not going to post it here. My advice is to just look for another approximation for $\sin$ rather than Taylor series.
This is just too long for the comments so don't bother up/downvoting this post. It's only an answer if the F function is sufficiently nicer than the amplitude function.
A: Expand $-\sin(y)$ to $-y+y^{3}/6$. Assume that the solution can be presented in the form
$$
y(x) = \sum_{i = 0}^{\infty} c_{i}x^{i}
$$
Then by the Cauchy product,
$$
y(x)^2 = \sum_{i = 0}^{\infty} \left(\sum_{j = 0}^{i} c_{j}c_{i-j}\right)x^{i} =: \sum_{i=0}^{\infty} 6b_{i}x^{i}.
$$
Then 
\begin{align}
y^{3}/6 - y 
&= y\left(y^{2}/6 -1 \right) = \sum_{j=0}^{\infty}c_{j}x^{j} \left(\sum_{i=0}^{\infty} b_{i}x^{i} -1 \right)\\
&= \sum_{i=0}^{\infty} \left(\sum_{j=0}^{i} c_{j}b_{i-j}  \right)x^{i}  - \sum_{i=0}^{\infty} c_{i}x^{i} \\
&= \sum_{i=0}^{\infty} \left[\sum_{j=0}^{i} c_{j}b_{i-j}  - c_{i}  \right]x^{i}
\end{align}
Since 
$$
y''(x) = \sum_{i=0}^{\infty} (i+1)(i+2)c_{i+2}x^{i}
$$
so that equating coefficients of like powers of $x$ yields
\begin{align}
c_{i+2} 
&= \frac{1}{(i+1)(i+2)}\left[\sum_{j=0}^{i} c_{j}b_{i-j}  - c_{i}  \right]  \\
&= \frac{1}{(i+1)(i+2)}\left[\sum_{j=0}^{i} \sum_{\ell=0}^{i-j}\frac{1}{6}c_{j}c_{\ell}c_{i-j-\ell}  - c_{i}  \right]  \\
\end{align}
It's expressions like these that make people love Runge-Kutta. But in any case, 
\begin{align}
c_{2} &= -\frac{1}{2}c_{0} && c_{3} = \frac{c_{1}}{6}\left(\frac{1}{2}c_{0}^{2} -1 \right)
\end{align}
and presumably you can program the recurrence relation to get as many terms as you'd like.
But the best solution is just to type DSolve[D[y[x],{x,2}] == -Sin[y], {y,x}] into Wolfram alpha and work with the result.
