Unit circle and right angle problem?

• Dec 20th 2010, 11:38 AM
homeylova223
Unit circle and right angle problem?
if cot(-180 degrees)
would you add 360 to get positive 180
then do cot (180) to get 1/0 or undefined

my second problem is if tan= 1.3
what is the value of (cot-angle)
I am not sure how to do this.

• Dec 20th 2010, 12:00 PM
homeylova223
Also I am trying to find the tan of 150 degrees
so I have to dive (1/2)/(- square root 3/2) can anyone also show me how to do this.
• Dec 20th 2010, 12:15 PM
Quote:

Originally Posted by homeylova223
if cot(-180 degrees)
would you add 360 to get positive 180
then do cot (180) to get 1/0 or undefined

my second problem is if tan= 1.3
what is the value of (cot-angle)
I am not sure how to do this.

A clockwise movement of $180^o$ brings us to the same position on the unit-circle

as a counterclockwise rotation of $180^o$

$sin\left[180^o\right]=sin\left[-180^o\right];\;\;\;cos\left[180^o\right]=cos\left[-180^o\right]$

Hence $cot\left[180^o\right]=cot\left[-180^o\right]$

$\displaystyle\ cotx=\frac{1}{tanx}=\frac{cosx}{sinx}$

For $x=-180^o,$ $sin\left[-180^o\right]=0$

hence the cotangent of that angle is undefined.

$\displaystyle\ tan\theta=1.3\Rightarrow\ cot\theta=\frac{1}{tan\theta}=\frac{1}{1.3}$

For cotangent, you only need know that it is the reciprocal of tangent.

$150^o=180^o-30^o$

$\displaystyle\ tan\left[150^o\right]=\frac{sin\left[150^o}{cos\left[150^o\right]}$

In the unit-circle, $sin\theta=sin\left[180^o-\theta\right]$

and $cos\theta=-cos\left[180^o-\theta\right]$

Therefore

$\displaystyle\frac{sin\left[150^o\right]}{cos\left[150^o\right]}=\frac{sin\left[30^o\right]}{-cos\left[30^o\right]}$

$=\displaystyle\frac{\left(\frac{1}{2}\right)}{-\left(\frac{\sqrt{3}}{2}\right)}=\frac{\left(\frac {1}{2}\right)}{\left(\frac{1}{2}\right)}\;\left[-\frac{1}{\sqrt{3}}\right]$