Thread: cot

**Petrus** · September 25th 2012, 05:38 AM

Calculate

. The answer may contain roots but not trigonometric functions.

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

**MarkFL** · September 25th 2012, 05:48 AM

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

**Petrus** · September 25th 2012, 06:01 AM

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

**MarkFL** · September 25th 2012, 06:11 AM

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

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

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

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

**Petrus** · September 25th 2012, 06:24 AM

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

**Petrus** · September 25th 2012, 06:29 AM

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

**MarkFL** · September 25th 2012, 06:33 AM

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

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:

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

**MarkFL** · September 25th 2012, 06:37 AM

Originally Posted by Petrus

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

You need to invert the result.

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