How do you find the value of #cot 240# using the double angle or half angle identity?

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Answer 1 · 2018-02-13T14:03:09

Double your angle. Double your fun.

Since #240^o# is one of the standard angles, no one would ever do this. However, it can be done. First, let's look at tangent.

#tan2u = (2tanu)/(1-tan^2u)#

So

#tan240^o = (2tan120^o)/(1-tan^2(120^o))#

and

#tan120^o = (2tan60^o)/(1-tan^2(60^o))#

Since #tan60^o = sqrt3#,
we have

#tan120^o = (2sqrt3)/(1-3)#
#tan120^o = -sqrt3#

Therefore
#tan240^o = (-2sqrt3)/(1-3)#
#= (-2sqrt3)/(-2) = sqrt3#

Now #cot240^o = 1/sqrt3 = sqrt3/3#.

Lying in Quadrant III, #cot240^o = cot60^o#, so this is correct.