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

1 Answer
Vishwaksen Reddy V.
Oct 19, 2015

# 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)#

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