# Thread: Trig Problem

1. ## Trig Problem

Looking for someone to verify my work ....

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

Solution

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

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

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

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

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

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

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

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

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

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

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

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