The terminal side of #theta# lies on the line #2x-y=0# in quadrant III, how do you find the values of the six trigonometric functions by finding a point on the line?

2 Answers
Feb 8, 2017

Answer:

#{:(sin(theta)=-2/sqrt(3),color(white)("XX"),csc(theta)=-sqrt(3)/2), (cos(theta)=-1/sqrt(3),,sec(theta)=-sqrt(3)), (tan(theta)=2,,cot(theta)=1/2) :}#

Explanation:

Any relation of the form #Ax+By=C# has a slope of (i.e. a #tan#) #-A/B#
So #2x-y=0# has a slope (#tan#) of #2#.

Furthermore, we can see that #2x-y=0# passes through the origin,
so we have the situation in the image below for a point on the line in Q III:

enter image source here
The six trigonometric values can be determined by applying their definitions based on "adjacent", "opposite", and "hypotenuse" sides.

Apr 12, 2017

Answer:

The answer is:
#sin theta = -(2sqrt(5))/5#

#cos theta = - sqrt(5)/5#

Because #sqrt(2^2 + (-1)^2) = sqrt(5)#

Explanation:

Using the distance formula you arrive at the square root of 5. This answer is verified via a Larson Precalculus book in the answer section.

EDIT: How I did this was by picking a relivant point that I could work with. You could also use the unit circle equation of #x^2 + y^2 = 1# to find the answer as well. With 2x - y = 0 moving the y to the right you end up with 2x = y (y = 2x). your unit circle equation becomes #x^2 + (2x)^2 = 1# and you solve from there to end up with the answers above.