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How do you find the exact 6 trigonometric ratios for the angle x whose radian measure is #(5pi)/6#?

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May 5, 2015

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

#cos [(5pi)/6)] = cos (Pi - Pi/6) = -cos (pi/6) = (-sqr3)/2#

#tan ((5pi)/6) = -1/(sqr3) = -(sqr3)/3#

#cot [(5pi)/6] = -sqr3#

#sec [(5pi)/6] = 2#

#csc [(5pi)/6] = -2/(sqr3)#

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