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How do you evaluate #cot((11pi)/6)#?

1 Answer

Sep 25, 2016

#cot(11pi)/6 = -sqrt3#

Explanation:

#cot((11pi)/6)#

Recall the identity #cottheta=costheta/sintheta#

Using the unit circle, #cos((11pi)/6)=sqrt3/2# and #sin((11pi)/6)=-1/2#

#cot((11pi)/6)=frac{sqrt3/2}{-1/2}=sqrt3/2 * -2/1=-sqrt3#

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