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

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

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

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

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

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

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

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