How do you find all six trigonometric functions of #(5pi)/6#?

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Answer 1 · 2015-05-22T03:48:44

On the trig unit circle:

#sin ((5pi)/6) = sin (pi - pi/6) = sin (pi/6) = 1/2#

#cos ((5pi)/6) = cos (pi - pi/6) = -cos (pi/6) = (-sqrt3)/2#

#tan ((5pi)/6) = 1/(-sqrt3) = (-sqrt3)/3#

#cot ((5pi)/6) = 1/tan ((5pi)/6)# =

#sec = 1/cos ((5pi)/6)#

#csc = 1/sin ((5pi)/6) =#