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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HSBC244 Share
Dec 21, 2016

You have to start by finding a point that lies on this line. I think the simplest would be #(-6, -1)# (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.

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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#

#(-1)^2 + (-6)^2= c^2#

#1 + 36 = c^2#

#c^2 = 37#

#c = +-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.

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

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

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

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

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

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

Hopefully this helps!

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