How do you use the half angle identify to find the exact value of #tan165^circ#?

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Answer 1 · 2016-12-11T09:33:58

#tan165^o=-2+sqrt3#

As #tan(180^o-A)=-tanA#, #tan165^o=tan(180^o-15^o)=-tan15^o#

For #tan15^o#, let us use half angle identity #tanA=(2tan(A/2))/(1-tan^2(A/2))#.

Using this if #A=30^o# and #tan30^o=1/sqrt3#, we have

#(2tanA)/(1-tan^2A)=1/sqrt3#, where #A=15^o#

or #1-tan^2A=2sqrt3tanA#

or #tan^2A+2sqrt3tanA-1=0#

and using quadratic formula

#tanA=(-2sqrt3+-sqrt((2sqrt3)^2-4×1×(-1)))/2#

or #tanA=(-2sqrt3+-sqrt(12+4))/2#

= #(-2sqrt3+-4)/2#

= #2-sqrt3# or #-2-sqrt3#

But #tan15^o# cannot be negative. Hence, #tan15^o=2-sqrt3# and

#tan165^o=-2+sqrt3#