How do you use half angle formula to find #tan (5pi)/12#?

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Answer 1 · 2015-04-28T01:13:36

#tan(5pi) -= tan(pi) = 0#
and there is no point in using the half angle formula to evaluate
#(tan(5pi))/12#

Therefore, let's assume that you really want to evaluate
#tan((5pi)/12)#

Half angle formula (for tan):
#tan^2(theta/2) = (1-cos(theta))/(1+cos(theta))#

#tan^2((5pi)/12) = tan(((5pi)/6)/2)#

#=(1-cos(pi/6))/(1+cos(pi/6))#

#= (1+sqrt(3)/2)/(1-sqrt(3)/2)#

#= (1+sqrt(3)/2)^2/(1-3/4)#

#= 4(1+sqrt(3)/2)^2#

That is
#tan^2((5pi)/12) = (2(1+sqrt(3)/2))^2#

Since #(5pi)/12# is in Quadrant 1, the tan is positive
and
#tan((5pi)/12) = 2+sqrt(3)#