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..

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- Jan 9th 2010, 06:31 PMPunchSimplifying
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.. - Jan 9th 2010, 06:46 PMProve It
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 $\displaystyle \sqrt{2 + \sqrt{2 + 2\cos{x}}} = \sqrt{2 + 2\cos{\frac{x}{2}}}$

$\displaystyle = \sqrt{2\left(1 + \cos{\frac{x}{2}}\right)}$

$\displaystyle = \sqrt{\frac{4\left(1 + \cos{\frac{x}{2}}\right)}{2}}$

$\displaystyle = 2\sqrt{\frac{1 + \cos{\frac{x}{2}}}{2}}$

$\displaystyle = 2\cos{\frac{x}{4}}$. - Jan 9th 2010, 06:49 PMPunch
Erm, sorry but the answer in the textbook is $\displaystyle 2sin\frac{x}{4}$

- Jan 9th 2010, 06:56 PMProve It
Then the original question should have been

$\displaystyle \sqrt{2 - \sqrt{2 + 2\cos{x}}}$. - Jan 9th 2010, 07:03 PMPunch
- Jan 9th 2010, 07:06 PMPunch
And if the answer provided in the book is wrong, i would like to ask if the step to solving such questions with a square root in another square root is to solve the square roots one by one before combining them all?

- Jan 9th 2010, 07:10 PMProve It
- Jan 9th 2010, 07:12 PMPunch
i think that this question is the toughest question i have met in this chapter though it looks easy