How do you use the half-angle identity to find the exact value of cot (11pi/12)?

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Answer 1 · 2015-07-14T14:41:49

Find #cot ((11pi)/12)#

Call #tan ((11pi)/12) = t#

tan #(22pi)/12 = tan ((10pi)/12 + pi) = tan ((5pi)/6) = - 1/(sqrt3)#

Recall #tan 2a = (2tan a)/(1 - tan^2 a)#

#tan (22pi)/12 = -1/(sqrt3) = (2t)/(1 - t^2) =#

#t^2 - 2sqrt3t - 1 = 0#

#D = d^2 = 12 + 4 = 16 --> d = +- 4#

#t = sqrt3 +- 2# -> #t = 2 +-sqrt3 #

#cot ((11pi)/12) = 1/t = 1/(2 +- sqrt3)#