Differential equation of a pendulum 
Consider the nonlinear differential equation of the pendulum $$\frac{d^2\theta}{dt^2}+\sin \theta=0$$ with $\theta(0)=\frac{\pi}3$ and $\theta'(0)=0$. Using the series method, find the first four nonzero terms of the solution.

Here is what I found from Maple so far:

Text-only (for the series solution): 
\begin{align}
\theta(t)&=\frac{\pi}3-\frac 12\sin\left(\frac{\pi}3\right)t^2+\frac 1{24}\sin\left(\frac{\pi}3\right)\cos\left(\frac{\pi}3\right)t^4+O(t^6)\\
&=\frac{\pi}3-\frac{\sqrt{3}}4t^2+\frac{\sqrt{3}}{96}t^4+O(t^6)
\end{align}
But how can one find this solution by hand, using the series method?
 A: Expanding on @Did's comment:
The initial condition $θ′(0)=0$ implies that the solution $t↦θ(t)$ is even. Express the series solution as $\theta(t)=c_0+u(t)$, with $u(t):=c_2t^2+c_4t^4+c_6t^6+O(t^8)$. The initial condition $\theta(0)=\frac{\pi}3$ shows that $$c_0=\frac{\pi}3.$$ Thus, $\theta(t)=\frac{\pi}3+u(t)$. This also means 
\begin{align}
\sin(\theta(t))&=\sin\left(\frac{\pi}3+u(t)\right) \\
&=\sin \left(\frac{\pi}3\right) \cos (u(t))+\cos \left(\frac{\pi}3\right) \sin( u(t)) \\
&= \frac{\sqrt{3}}2 \cos (u(t))+\frac 12 \sin (u(t))
\end{align}
Now, $$\frac{\sqrt{3}}2\cos(u(t))=\frac{\sqrt{3}}2-\frac{\sqrt{3}}4 c_2 t^4 + O(t^6)$$ and $$\frac 12\sin(u(t))=\frac 12c_2t^2+\frac 12c_4t^4+O(t^6).$$
Therefore, 
$$\sin(\theta(t))=\frac{\sqrt{3}}2+\frac 12c_2t^2+\left(-\frac{\sqrt{3}}4c_2+\frac 12c_4\right)t^4+O(t^6)$$
Now, the ODE $\theta''(t)=-\sin(\theta(t))$, written in series expansion form, becomes $$2c_2+12c_4t^2+30c_6t^4+O(t^6)=-\left[\frac{\sqrt{3}}2+\frac 12c_2t^2-\left(\frac 12c_4-\frac{\sqrt{3}}4 c_2\right)t^4+O(t^6) \right]$$
Equating the coefficients gives $$c_2=-\frac{\sqrt{3}}4,\quad c_4=\frac{\sqrt{3}}{96},\quad c_6=\frac {\sqrt{3}}{5760}+\frac 1{160}.$$
