1. ## cot

Calculate . The answer may contain roots but not trigonometric functions.

im really bad on the geometry so any1 give me tips?

2. ## Re: cot

Hint: $\displaystyle \cot(\theta+k\pi)=\cot(\theta)$ where $\displaystyle k\in\mathbb{Z}$.

3. ## 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

4. ## Re: cot

$\displaystyle -\frac{31}{6}=-5\frac{1}{6}$ so you have:

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

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

$\displaystyle 0<\frac{5}{6}\pi<\pi$

5. ## Re: cot

ty again! but how can ik what cot 5pi/6 is on test

6. ## Re: cot

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

7. ## Re: cot

$\displaystyle \cot(x)=\frac{\cos(x)}{\sin(x)}$

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

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

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

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

$\displaystyle \cot\left(-\frac{31\pi}{6} \right)=-\cot\left(\frac{31\pi}{6}-5\pi \right)=-\cot\left(\frac{\pi}{6} \right)$

8. ## Re: cot

Originally Posted by Petrus
nvm solved it
cot (5π/6) = 1 / tan (5π/6) = - 1/√3
You need to invert the result.