How do you find the exact value of the five remaining trigonometric function given sintheta=1/3 and the angle is in standard position in quadrant II?

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Apr 1, 2017

$\sin \left(\theta\right) = \frac{1}{3}$
$\cos \left(\theta\right) = - \frac{2 \sqrt{2}}{3}$
$\tan \left(\theta\right) = - \frac{\sqrt{2}}{4}$
$\csc \left(\theta\right) = 3$
$\sec \left(\theta\right) = - \frac{3 \sqrt{2}}{4}$
$\cot \left(\theta\right) = - 2 \sqrt{2}$

Explanation:

First, $\csc \left(\theta\right) = \frac{1}{\sin} \left(\theta\right) = \frac{1}{\frac{1}{3}} = 3$.

We know that ${\sin}^{2} \left(\theta\right) + {\cos}^{2} \left(\theta\right) = 1$, or $\cos \left(\theta\right) = \pm \sqrt{1 - {\sin}^{2} \left(\theta\right)} = \pm \sqrt{1 - {\left(\frac{1}{3}\right)}^{2}} = \pm \frac{2 \sqrt{2}}{3}$. Now, the angle is in quadrant 2. This means that its cosine value is negative. Thus, $\cos \left(\theta\right) = - \frac{2 \sqrt{2}}{3}$.

Then, $\sec \left(\theta\right) = \frac{1}{\cos} \left(\theta\right) = \frac{1}{- \frac{2 \sqrt{2}}{3}} = - \frac{3}{2 \sqrt{2}} = - \frac{3 \sqrt{2}}{4}$.

We know that $\tan \left(\theta\right) = \sin \frac{\theta}{\cos} \left(\theta\right) = \frac{\frac{1}{3}}{- \frac{2 \sqrt{2}}{3}} = - \frac{1}{2 \sqrt{2}} = - \frac{\sqrt{2}}{4}$.

Then, $\cot \left(\theta\right) = \frac{1}{\tan} \left(\theta\right) = \frac{1}{- \frac{\sqrt{2}}{4}} = - \frac{4}{\sqrt{2}} = - 2 \sqrt{2}$.

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