SOLUTION: Given sin(theta) = -3/7 and tan(theta) > 0, what are all 6 exact trig values and what is the corresponding quadrant and point on the circle?

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Question 387970: Given sin(theta) = -3/7 and tan(theta) > 0, what are all 6 exact trig values and what is the corresponding quadrant and point on the circle?
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
sin is negative in the 3rd and 4th quadrants. tan is positive in the 1st and 3rd quadrants. So for theta to have a negative sin and a positive tan, it must terminate in the 3rd quadrant. And in the third quadrant sin, cos, sec and csc are all negative while tan and cot are positive.

We can use the Pythagorean Theorem or cos^2(theta) = 1 - sin^2(theta) to find cos(theta):
cos^2(theta) = 1 - (-3/7)^2
cos^2(theta) = 1 - 9/49
cos^2(theta) = 40/49
Since cos is negative in the 3rd quadrant we will use only the negative square root:
cos(theta) =

csc(theta) = 1/sin(theta) = 1%2F%28-3%2F7%29+=+-7%2F3
sec(theta) = 1/cos(theta) =
tan(theta) - sin(theta)/cos(theta) =
cot(theta) = cos(theta)/sin(theta) = %28-2sqrt%2810%29%2F7%29%2F%28-3%2F7%29+=+2sqrt%2810%29%2F3

The "point on the circle" is (cos(theta), sin(theta) = (-2sqrt%2810%29%2F7, -3%2F7)