If $$y(h) = 1-2\sin^2(2\pi h), \quad f(y) = \frac{2}{1 + \sqrt{1-y^2}}$$ I would like to show that $$f(y(h)) = 2 - 4\sqrt{2}\pi + O(h^2).$$
Using the identity $$\sin^2{\theta} = \frac{1-2\cos(2\theta)}{2}$$ I was able to make the following simplifications. $$y(h) = 2\cos(4\pi h) \implies f(y(h)) = \frac{2}{1 + \sqrt{1- 4\cos^2(4\pi h)}}.$$
From here I am thinking we must use a Taylor expansion of cosine, the square root function, or both. Any hints on how to proceed?