Clever derivation of $\arcsin(x)$ Taylor series

Solution 1:

Here's a concise argument, but it uses $\sin^2 x+\cos^2x=1$, which isn't exactly obvious from the power series. Differentiating

$$x=\sum_{n=0}^\infty c_n\sin^nx$$

twice yields

$$ \begin{eqnarray} 0 &=& \sum_{n=0}^\infty c_n\left(n(n-1)\sin^{n-2}x\cos^2x-n\sin^nx\right) \\ &=& \sum_{n=0}^\infty c_n\left(n(n-1)\sin^{n-2}x-n^2\sin^nx\right)\;, \end{eqnarray} $$

which gives the recurrence

$$ \begin{eqnarray} c_{n+2} &=& \frac{n^2}{(n+1)(n+2)}c_n \\ &=& \frac{n}{n+2}\frac{n}{n+1}c_n\;. \end{eqnarray}$$

The factors of $n$ and $n+2$ cancel except for the final $n+2$, and with $c_0=0$ and $c_1=1$ this leads to

$$c_{2n+1}=\frac{(2n)!}{(2^nn!)^2(2n+1)}\;.$$

I got the idea for this here (item 4.3, third proof).