Simplifying

**Punch** · January 9th 2010, 06:31 PM

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..

**Prove It** · January 9th 2010, 06:46 PM

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

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

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

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

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

**Punch** · January 9th 2010, 06:49 PM

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

**Prove It** · January 9th 2010, 06:56 PM

Then the original question should have been

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

**Punch** · January 9th 2010, 07:03 PM

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

**Punch** · January 9th 2010, 07:06 PM

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?

**Prove It** · January 9th 2010, 07:10 PM

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?

Yes, start with the innermost square root and work out...

**Punch** · January 9th 2010, 07:12 PM

i think that this question is the toughest question i have met in this chapter though it looks easy

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