$\cos (2x) = \cos^2 x - \sin^2 x = 1-2\sin^2 x$ , assim, reescrevamos $y$ :$$y = \frac{1}{2} \cdot \frac{2+2^{1-2\sin^2 x}}{1+2^{2\sin^2 x}} = \frac{1+2^{-2\sin^2 x}}{1+2^{2\sin^2 x}}$$$$y = 2^{-2\sin^2 x} \cdot \underbrace{\frac{1+2^{2\sin^2 x}}{1+2^{2\sin^2 x}}}_{1}$$$$\boxed{y = 2^{-2\sin^2 x}}$$$$\text{Alternativa } \mathbb{(C)}$$