By the half-angle formula: $$\alpha\in\Big(\frac{3\pi}{2}, 2\pi\Big) \implies \sqrt{\frac{1}{2}+\frac{1}{2}\cdot \cos 2\alpha}=|\cos\alpha| = \cos\alpha$$ also $$\frac{\alpha}{2}\in\Big(\frac{3\pi}{4},\pi\Big)\implies\sqrt{\frac{1}{2}-\frac{1}{2}\cdot\cos\alpha}=\Big{|}\sin\frac{\alpha}{2}\Big{|}=\sin\frac{\alpha}{2}$$ Hence the final result is $\boxed{\sin\frac{\alpha}{2}}$.