# How do you find the values of the six trigonometric functions given sintheta=3/5 and theta lies in Quadrant II?

Feb 28, 2017

See below

#### Explanation:

$\sin \theta = \frac{3}{5}$ and $\theta$ is in Quadrant $\text{II}$

This means $\sin \left(180 - \theta\right) = \frac{3}{5}$

So in a right-angled triangle, the $\text{opp} = 3$ and the $\text{hyp} = 5$. Thus the $\text{adj} = \sqrt{{5}^{2} - {3}^{2}} = \sqrt{16} = 4$

Therefore, $\cos \left(180 - \theta\right) = \frac{4}{5}$

$\cos 180 \cos \theta + \sin 180 \sin \theta = \frac{4}{5}$

$- \cos \theta = \frac{4}{5}$

$\cos \theta = - \frac{4}{5}$

$\tan \theta = \sin \frac{\theta}{\cos} \theta = \frac{\frac{3}{5}}{- \frac{4}{5}} = - \frac{3}{4}$

$\csc \theta = \frac{1}{\sin} \theta = \frac{1}{\frac{3}{5}} = \frac{5}{3}$

$\sec \theta = \frac{1}{\cos} \theta = \frac{1}{- \frac{4}{5}} = - \frac{5}{4}$

$\cot \theta = \frac{1}{\tan} \theta = \frac{1}{- \frac{3}{4}} = - \frac{4}{3}$