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?
I see that @Mike post contains valuable points but I'd like you to note some other ones:


  If $~~\sin(t)=\sin(s)$ then $t=2k\pi+s, ~~\text{or}~~~~t=2k\pi+\pi-s$.


Now consider that that angle is in $[\pi/2,\pi]$ so we have one choice here. Here $s=\pi/6$ and indeed $k=0$ gives it to us. That is $\pi-\pi/6=5\pi/6$. Since we'd like the angle to be in second quadrant, so the sing of $\cos$ is in fact negative so $$\cos(5\pi/6)=\sqrt{1-\sin^2(5\pi/6)}=\sqrt{1-1/4}=|\frac{\sqrt{3}}{2}|=-\sqrt{3}/2.$$
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, $\cos\theta = \cos \frac {5\pi}{6} = \cos (\pi - \frac{\pi}{6}) = … = - \cos \frac {\pi}{6} = -\frac {\sqrt 3}{2}$