# The terminal side of theta lies on the line y=1/3x in quadrant III, how do you find the values of the six trigonometric functions by finding a point on the line?

Mar 11, 2017

We're in the third quadrant, where both the x-axis and the y-axis is negative.

I would recommend looking for the first coordinate on the line after the origin where both x and y values are integers. This would be $\left(- 3 , - 1\right)$.

We now have enough information to determine $\tan \theta$ and $\cot \theta$.

$\tan \theta = \text{opposite"/"adjacent} = - \frac{1}{- 3} = \frac{1}{3}$

cottheta = 1/tantheta = 1/("opposite"/"adjacent") = "adjacent"/"opposite" = (-3)/(-1) = 3

We must find the hypotenuse of the triangle to find the other ratios.

${\left(- 3\right)}^{2} + {\left(- 1\right)}^{2} = {h}^{2}$

$9 + 1 = {h}^{2}$

$h = \sqrt{10}$

We can now apply the definitions of the other ratios to solve.

$\sin \theta = \text{opposite"/"hypotenuse} = - \frac{1}{\sqrt{10}} = \frac{- \sqrt{10}}{10}$

$\cos \theta = \text{adjacent"/"hypotenuse} = - \frac{3}{\sqrt{10}} = \frac{- 3 \sqrt{10}}{10}$

$\sec \theta = \text{hypotenuse"/"adjacent} = \frac{\sqrt{10}}{-} 3 = - \frac{\sqrt{10}}{3}$

$\csc \theta = \text{hypotenuse"/"opposite} = \frac{\sqrt{10}}{-} 1 = - \sqrt{10}$

Hopefully this helps!