How do you use the half angle identify to find the exact value of #tan((5pi)/12)#?

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Answer 1 · 2017-06-16T13:34:35

The answer is #=sqrt((2+sqrt3)/(2-sqrt3))#

We apply

#cos2t=2cos^2t-1#

#cost=sqrt((1+cos2t)/2)#

#cos2t=1-2sin^2t#

#sint=sqrt((1-cos2t)/2)#

#tant=sint/cost=sqrt((1-cos2t)/(1+cos2t))#

Here, we have #t=5/12pi#

#2t=10/12pi=5/6pi#

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

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

#tan(5/12pi)=sqrt((1+sqrt3/2)/(1-sqrt3/2))#

#=sqrt((2+sqrt3)/(2-sqrt3))#