If $\sin B = −1/2$ with $B$ in QIII, find $\cos B/2$

Question

For the following, assume that all the given angles are in simplest form, so that if A is in QIV you may assume that 270° < A < 360°.

If $\sin B = −1/2$ with B in QIII, find $\cos B/2$

Here's my attempt. Am I correct? $ -sqrt((-sqrt(3)+2)/4)$

$ -\sqrt{\frac{-\sqrt3+2}4}$

http://www.wolframalpha.com/input/?i=-sqrt%28%28-sqrt%283%29%2B2%29%2F4%29

Answer 1

So you have the following:

$$ \cos(2x) = \cos^2(x) - \sin^2(x) = \cos^2(x) - (1 - \cos^2(x)) = 2\cos^2(x) - 1 \\ \cos(x) = \pm\sqrt{\frac{1 + \cos(2x)}{2}} \rightarrow \cos\left(\frac{x}{2}\right) = \pm\sqrt{\frac{1 + \cos(x)}{2}} $$

You can find $\cos(B)$ by drawing a picture:

This gives:

$$ \cos\left(\frac{B}{2}\right) = \pm\sqrt{\frac{1 - \frac{\sqrt{3}}{2}}{2}} = \pm\frac{\sqrt{2 - \sqrt{3}}}{2} $$

So should it be plus or minus? If $\frac{B}{2}$ is in quadrant I, it's positive and if $\frac{B}{2}$ is in quadrant II, it should be negative. It should be easy to see that $\frac{B}{2}$ must be in quadrant II because if we say we have $0 \leq b \leq \frac{\pi}{2}$, then $B = \pi + b$ and $\frac{B}{2} = \frac{\pi}{2} + \frac{b}{2}$. This puts $\frac{B}{2}$ squarely in the second quadrant (you're adding $90^\circ$ to half of something less than $90^\circ$). This means:

$$ \cos\left(\frac{B}{2}\right) = -\frac{\sqrt{2 - \sqrt{3}}}{2} $$

Answer 2

If $B$ is in QIII, $\sin B$ and $\cos B$ are both negative. So \begin{align} \cos B & = -\sqrt{1-\sin^2 B} & \text{due the identity $\sin^2 B + \cos^2 B = 1$ } \\ &=-\sqrt{1-\frac{1}{4}} & \\ &=-\frac{\sqrt{3}}{2} \end{align} Also, $\frac{B}{2}$ is in QII, so $\cos \frac{B}{2}$ is negative, and using the identity $\cos^2\frac{t}{2}=\frac{1}{2}(1-\cos t)$ we get $\cos\frac{B}{2}=-\sqrt{\frac{1}{2}[1-(-\frac{\sqrt{3}}{2})]}=-\frac{1}{2}\sqrt{2-\sqrt{3}}$..