Question 387970
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.<br>
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) = {{{-sqrt(40/49) = -sqrt(40)/sqrt(49) = -(sqrt(4)*sqrt(10))/sqrt(49) = -2sqrt(10)/7}}}<br>
csc(theta) = 1/sin(theta) = {{{1/(-3/7) = -7/3}}}
sec(theta) = 1/cos(theta) = {{{1/(-2sqrt(10)/7) = -7/2sqrt(10) = (-7/2sqrt(10))(sqrt(10)/sqrt(10)) = -7sqrt(10)/20}}}
tan(theta) - sin(theta)/cos(theta) = {{{(-3/7)/(-2sqrt(10)/7) = 3/2sqrt(10) = (3/2sqrt(10))(sqrt(10)/sqrt(10)) = 3sqrt(10)/20}}}
cot(theta) = cos(theta)/sin(theta) = {{{(-2sqrt(10)/7)/(-3/7) = 2sqrt(10)/3}}}<br>
The "point on the circle" is (cos(theta), sin(theta) = ({{{-2sqrt(10)/7}}}, {{{-3/7}}})