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

1 Answer

cos theta = sqrt(3)/2cosθ=32
sec theta = 2/sqrt(3)secθ=23
sin theta = -1/2sinθ=12
cosec theta = -2cosecθ=2
tan theta = -1/sqrt(3)tanθ=13
cot theta = -sqrt(3)cotθ=3

Explanation:

11pi/6 = 2pi - pi/611π6=2ππ6
lies in the 4th quadrant
cos and sec are positive and remaining are negative
cos 2pi - theta = cos thetacos2πθ=cosθ
so cos (2pi - pi/6) = cos (pi/6)cos(2ππ6)=cos(π6)
cos theta = sqrt(3)/2cosθ=32
so sec theta = 2/sqrt(3)secθ=23
sin theta = -1/2sinθ=12
cosec theta = -2cosecθ=2
tan theta = -1/sqrt(3)tanθ=13
cot theta = -sqrt(3)cotθ=3