Prove the formula $$\sin^2 \frac{x}{2} = \sum_{n=1}^\infty C_{n-1} \left ( \frac{\sin x}{2} \right )^{2n} $$ where $C_n = \displaystyle \frac{1}{n+1}{{2n}\choose{n}}$.
I tried to use the Taylor series of sine function to obtain an infinite sum, but the exponent $2n$ is difficult to deal with.
Hint. Note that Catalan numbers $(C_n)_n$ have a nice generating function: for $|4z|\leq 1$, $$\sum_{n = 0}^{\infty} C_n z^n = \frac{1-\sqrt{1-4z}}{2z}.$$ Now note that $$\sum_{n=1}^\infty C_{n-1} \left ( \frac{\sin x}{2} \right )^{2n}=\frac{\sin^2 x}{4}\sum_{n=0}^\infty C_{n} \left (\frac{\sin^2 x}{4} \right )^{n}.$$ Can you take it from here?