How do you evaluate #Cot (π/3) - cos(π/6)#?

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Answer 1 · 2016-06-14T08:21:21

It is #-1/(2sqrt(3))#.

The definition of #cot(x)# is

#cot(x)=cos(x)/sin(x)#. So we have

#cot(pi/3)-cos(pi/6)#

#=cos(pi/3)/sin(pi/3)-cos(pi/6)#

Here we can substitute the values

#cos(pi/3)=1/2#

#sin(pi/3)=sqrt(3)/2#

#cos(pi/6)=sqrt(3)/2#

then we have

#cos(pi/3)/sin(pi/3)-cos(pi/6)#

#=(1/2)/(sqrt(3)/2)-sqrt(3)/2#

#=1/2*2/sqrt(3)-sqrt(3)/2#

#=1/sqrt(3)-sqrt(3)/2#

#=(2-3)/(2sqrt(3))#

#=-1/(2sqrt(3))#.