1. ## Simplifying

If $270^\circ < x < 360^\circ$, simplify $\sqrt{2+\sqrt{2+2cosx}}$

I couldn't do anything else after getting rid of the square roots..

2. Originally Posted by Punch
If $270^\circ < x < 360^\circ$, simplify $\sqrt{2+\sqrt{2+2cosx}}$

I couldn't do anything else after getting rid of the square roots..
Use the half angle identity for cosine:

$\cos{\frac{\theta}{2}} = \sqrt{\frac{1 + \cos{\theta}}{2}}$.

$\sqrt{2 + 2\cos{x}} = \sqrt{2(1 + \cos{x})}$

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

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

$= 2\cos{\frac{x}{2}}$.

Therefore $\sqrt{2 + \sqrt{2 + 2\cos{x}}} = \sqrt{2 + 2\cos{\frac{x}{2}}}$

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

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

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

$= 2\cos{\frac{x}{4}}$.

3. Erm, sorry but the answer in the textbook is $2sin\frac{x}{4}$

4. Then the original question should have been

$\sqrt{2 - \sqrt{2 + 2\cos{x}}}$.

5. Originally Posted by Prove It
Then the original question should have been

$\sqrt{2 - \sqrt{2 + 2\cos{x}}}$.
But i just doubled check both the answer and the question, it is a + for the question

6. 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?

7. Originally Posted by Punch
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?