\displaystyle \sin^2z=1-\cos^2z\\
\cos 2z=\cos^2z-\sin^2z=2\cos^2-1 \Longrightarrow \cos^2 z = \frac{1+\cos 2z}{2}\\
\Longrightarrow \sin^2 z=\frac{1}{2}-\frac{1}{2}\sum_{n\in\mathbb{N}_0}\left( -1\right)^n\frac{\left( 2z\right)^{2n}}{\left( 2n\right) !}=\frac{1}{2}\sum_{n\in\mathbb{N}}\left( -1\right)^{n-1}\frac{2^{2n}}{\left( 2n\right) !}z^{2n}