# How do I simplify a radical within a radical in this half-angle problem?

I don't understand how to simplify the following radicals and arrive at the final answer below. I can make it to this point:

$$\sin\left(-\frac{3\pi}{8}\right)=\pm\sqrt{1+\frac{\sqrt2}{2}\over2}$$

$$\sin\left(-\frac{3\pi}{8}\right)=-\frac{\sqrt{2+\sqrt{2}}}{2}$$

I've filled a couple pages and tried finding a good answer on how to do this simplification, but without success. Any help is gratefully received!

edit: removed the square over the denominator 2 in the final answer.

• Are you very sure that this is supposed to be the correct answer? Unless I'm missing something profound I don't think that those two are equal... – Flint72 Apr 10 '14 at 3:48
• I made a mistake: the 2 in the answer's denominator should not be squared. Thank you for pointing that out! – Jason S. Apr 10 '14 at 3:54
• No problem, happy to help! :) So can you edit your original post to correct for this mistake please, so that we see what it should be. Thanks! – Flint72 Apr 10 '14 at 3:57

First, $\;-\frac\pi2<-\frac{3\pi}8< 0\;$ , so we're in the fourth quadrant and thus sine is negative here. Second:
$$\sqrt{\frac{1+\frac{\sqrt2}{2}}{2}}=\sqrt{\frac{\frac{2+\sqrt2}2}{2}}=\sqrt{\frac{2+\sqrt2}4}=\frac{\sqrt{2+\sqrt2}}2$$
• Thank you for such a clear explanation! It appears I was trying to rationalize the denominator at your 2nd step, when I should have multiplied the complex fraction (or the numerator) under the radical by $\frac{1}{2}$ and then taken the square root of 4 in the denominator to simplify. – Jason S. Apr 10 '14 at 14:19
• Correction: I was supposed to multiply $\sqrt{\frac{2+\sqrt{2}}{2}\over{2}}$ by 2 in the numerator and denominator to eliminate the complex fraction, giving me a new denominator of 4. – Jason S. Apr 11 '14 at 0:49