If $y_n=\sin^2(\frac{\pi}{2}x_n)$, how do you express the $y_{n+1}$
I attempt to transform $\sin^2(\frac{\pi}{2}x_n)$ into $\frac{1}{2} ({1-\cos(\pi x_n))}$ by using double-angle formula, but I have no idea what to do from this. Any advice will be appericiated.
$$x_{n+1} = \begin{cases} 2x_n & 0<x_n< 1/2\\ 2(1-x_n) & 1/2\leq x_n\leq 1 \\ \end{cases} $$
If $0<x_n <1/2$ then $\sin(\frac{\pi}{2}x_{n+1})=\sin(\frac{\pi}{2}2x_n)= \sin( \pi x_n).$.
If $ 1/2 \le x_n \le 1$, then $x_{n+1}=2(1-x_n)$. Now use the addition theorem for $\sin$ to get again
$\sin(\frac{\pi}{2}x_{n+1})=\sin( \pi x_n).$
Hence: $y_{n+1}=\sin^2( \pi x_n).$