How do you find the exact value of all 6 trigonometric functions for the angle 11*pi / 6?

1 Answer

cos theta = sqrt(3)/2
sec theta = 2/sqrt(3)
sin theta = -1/2
cosec theta = -2
tan theta = -1/sqrt(3)
cot theta = -sqrt(3)

Explanation:

11pi/6 = 2pi - pi/6
lies in the 4th quadrant
cos and sec are positive and remaining are negative
cos 2pi - theta = cos theta
so cos (2pi - pi/6) = cos (pi/6)
cos theta = sqrt(3)/2
so sec theta = 2/sqrt(3)
sin theta = -1/2
cosec theta = -2
tan theta = -1/sqrt(3)
cot theta = -sqrt(3)