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# The terminal side of θ lies on a given line in the specified quadrant. Find the values of the six trigonometric functions of θ by finding a point on the line? Line y = 1/6x Quad: III

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Noah G Share
Dec 21, 2016

You have to start by finding a point that lies on this line. I think the simplest would be $\left(- 6 , - 1\right)$ (since we are in quadrant III, both the x and y axis have to have negative values).

We now draw an imaginary triangle, as shown in the following diagram.

We can now clearly see that our side opposite $\theta$ measures $1$ unit and our side adjacent $\theta$ measures $6$ units.

We must finally find the hypotenuse prior to determining the ratios. By pythagorean theorem:

${a}^{2} + {b}^{2} = {c}^{2}$

${\left(- 1\right)}^{2} + {\left(- 6\right)}^{2} = {c}^{2}$

$1 + 36 = {c}^{2}$

${c}^{2} = 37$

$c = \pm \sqrt{37}$

However, the hypotenuse can never have a negative length, so we can only accept the positive solution. We can now find our ratios.

$\sin \theta = \text{opposite"/"hypotenuse} = - \frac{1}{\sqrt{37}} = - \frac{\sqrt{37}}{37}$

csctheta = 1/sintheta = 1/("opposite"/"hypotenuse") = "hypotenuse"/"opposite" = sqrt(37)/(-1) = -sqrt(37)

$\cos \theta = \text{adjacent"/"hypotenuse} = - \frac{6}{\sqrt{37}} = \frac{- 6 \sqrt{37}}{37}$

sectheta = 1/costheta = 1/("adjacent"/"hypotenuse") = "hypotenuse"/"adjacent" = sqrt(37)/(-6) = -sqrt(37)/6

$\tan \theta = \text{opposite"/"adjacent} = - \frac{1}{- 6} = \frac{1}{6}$

cottheta = 1/tantheta = 1/("opposite"/"adjacent") = "adjacent"/"opposite" = -6/(-1) = 6

Hopefully this helps!

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