<s>[INST] Use the range for $\theta$ to determine the indicated function value Problem:

$$\sin\theta = \frac {1}{2}; \frac {\pi}{2} \leq \theta \leq \pi, \text{find} \cos\theta$$

So I have that $\sin\theta = \frac {1}{2}$ and that the range for $\theta$ is $\frac {\pi}{2} \leq \theta \leq \pi$.
I'm assuming I need to find the $\theta$ for which $\sin\theta = \frac {1}{2}$ so I can enter that value into the cos function but I'm not sure how to go about it.
I know $\sin\frac {\pi}{2} = 1$ and $\sin\pi$ = 0. So I assumed.. maybe $\sin\frac{3\pi}{4}$ would be $\frac{1}{2}$ but I don't think that's correct. How would I go about solving this?
 [/INST] Solving $\sin\theta = \frac {1}{2}$, we get $\theta = \frac {\pi}{3}$ or $\frac {2\pi}{3}$. 
Edit: my mistake and should be $\theta = \frac {\pi}{6}$ or $\frac {5\pi}{6}$
Note that only the latter fits the given range.
Thus, $\cos\theta = \cos \frac {5\pi}{6} = \cos (\pi - \frac{\pi}{6}) = … = - \cos \frac {\pi}{6} = -\frac {\sqrt 3}{2}$
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