If $\displaystyle 270^\circ < x < 360^\circ$, simplify $\displaystyle \sqrt{2+\sqrt{2+2cosx}}$
I couldn't do anything else after getting rid of the square roots..
Use the half angle identity for cosine:
$\displaystyle \cos{\frac{\theta}{2}} = \sqrt{\frac{1 + \cos{\theta}}{2}}$.
$\displaystyle \sqrt{2 + 2\cos{x}} = \sqrt{2(1 + \cos{x})}$
$\displaystyle = \sqrt{\frac{4(1 + \cos{x})}{2}}$
$\displaystyle = 2\sqrt{\frac{1 + \cos{x}}{2}}$
$\displaystyle = 2\cos{\frac{x}{2}}$.
Therefore $\displaystyle \sqrt{2 + \sqrt{2 + 2\cos{x}}} = \sqrt{2 + 2\cos{\frac{x}{2}}}$
$\displaystyle = \sqrt{2\left(1 + \cos{\frac{x}{2}}\right)}$
$\displaystyle = \sqrt{\frac{4\left(1 + \cos{\frac{x}{2}}\right)}{2}}$
$\displaystyle = 2\sqrt{\frac{1 + \cos{\frac{x}{2}}}{2}}$
$\displaystyle = 2\cos{\frac{x}{4}}$.