# Simplifying

• Jan 9th 2010, 07:31 PM
Punch
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..
• Jan 9th 2010, 07:46 PM
Prove It
Quote:

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

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

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

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

$= 2\cos{\frac{x}{4}}$.
• Jan 9th 2010, 07:49 PM
Punch
Erm, sorry but the answer in the textbook is $2sin\frac{x}{4}$
• Jan 9th 2010, 07:56 PM
Prove It
Then the original question should have been

$\sqrt{2 - \sqrt{2 + 2\cos{x}}}$.
• Jan 9th 2010, 08:03 PM
Punch
Quote:

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
• Jan 9th 2010, 08:06 PM
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?
• Jan 9th 2010, 08:10 PM
Prove It
Quote:

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?