How do you evaluate #cot ((7pi)/6)#?

1 Answer
Konstantinos Michailidis
Jul 9, 2016

It is

#cot((pi+6pi)/6)=cot(pi+pi/6)=cot(pi/6)=sqrt3#

Another way is that

#cot(pi+pi/6)=1/(tan(pi+pi/6))#

Use the #tan(a + b) = [tan a + tan b]/[1 - (tan a)(tan b)] #

so you get:

#tan(pi + pi/6) = [tanpi + tanpi/6]/[1 - (tanpi)(tanpi/6)] = [ 0 + (sqrt3)/3] / [1 - (0)((sqrt3)/3)] = [(sqrt3)/3] / [1 - 0] = [(sqrt3)/3] / 1 = [sqrt3] / 3 #

Hence #cot(pi+pi/6)=1/(sqrt3/3)=sqrt3#

Related questions
See all questions in Trigonometric Functions of Any Angle
Impact of this question
20384 views around the world
You can reuse this answer
Creative Commons License