1. ## Trig Problem

Looking for someone to verify my work ....

If $\sin\theta=\frac{1}{3}$ and $\frac{\pi}{2}<\theta<\pi$ then what is $\cos\theta?$

Solution

$\sin^2\theta+\cos^2\theta=1$

$\frac{1}{3}^2+\cos^2\theta=1$

$\cos^2\theta=1-\frac{1}{9}$

$\cos^2\theta=\frac{8}{9}$

$\cos\theta=\frac{\sqrt{8}}{3}$

$\cos\theta=\frac{2\sqrt{2}}{3}$

Now because $\frac{\pi}{2}<\theta<\pi$ then my answer should read $\cos\theta=\frac{-2\sqrt{2}}{3}$

I assume that I have completed everything properly.

2. Yes my good sir, you are correct.

And just to help you along, here's an alternative method to do the question.

If $sin\theta=\frac{1}{3}$, then you may construct the right-angled $\Delta$ with that ratio.

|\
| \
1 | \ 3
| \
|____\
$\frac{2\sqrt{2}}{3}$

So $cos\theta=\frac{2\sqrt{2}}{3}$}.

But since the desired range is $\frac{\pi}{2}<\theta<\pi$
use the identity $cos(180-\theta)=-cos\theta$, to get the desired result

Forgive my very shoddy triangle please. It's almost 11p.m.