# cot

• September 25th 2012, 05:38 AM
Petrus
cot
Calculate https://webwork.math.su.se/webwork2_...00c416d651.png. The answer may contain roots but not trigonometric functions.

im really bad on the geometry so any1 give me tips?
• September 25th 2012, 05:48 AM
MarkFL
Re: cot
Hint: $\cot(\theta+k\pi)=\cot(\theta)$ where $k\in\mathbb{Z}$.
• September 25th 2012, 06:01 AM
Petrus
Re: cot
i want the range 0 ≤ x < 2π, so i have to choose k so it will be on tat range so k= 4 if i got right,sorry kkkeybord broken
• September 25th 2012, 06:11 AM
MarkFL
Re: cot
$-\frac{31}{6}=-5\frac{1}{6}$ so you have:

$-6\pi<-\frac{31}{6}\pi<-5\pi$

$0<-\frac{31}{6}\pi+6\pi<\pi$

$0<\frac{5}{6}\pi<\pi$
• September 25th 2012, 06:24 AM
Petrus
Re: cot
ty again! but how can ik what cot 5pi/6 is on test
• September 25th 2012, 06:29 AM
Petrus
Re: cot
nvm solved it
cot (5π/6) = 1 / tan (5π/6) = - 1/√3
• September 25th 2012, 06:33 AM
MarkFL
Re: cot
$\cot(x)=\frac{\cos(x)}{\sin(x)}$

You should know (or be able to quickly find using the unit circle):

$\cos\left(\frac{5\pi}{6} \right)=-\frac{\sqrt{3}}{2}$

$\sin\left(\frac{5\pi}{6} \right)=\frac{1}{2}$

If you prefer to use a first quadrant angle, then you may write:

$\cot\left(-\frac{31\pi}{6} \right)=-\cot\left(\frac{31\pi}{6}-5\pi \right)=-\cot\left(\frac{\pi}{6} \right)$
• September 25th 2012, 06:37 AM
MarkFL
Re: cot
Quote:

Originally Posted by Petrus
nvm solved it
cot (5π/6) = 1 / tan (5π/6) = - 1/√3

You need to invert the result.