# unit circle help

• Sep 21st 2007, 09:38 PM
whitestrat
unit circle help
hi, i need a little bit of help with questions such as:

use a unit circle diagram to find all angles between 0deg and 360deg:

sine 0.5

cos 1/root2

etc...

i can usually do this as i have a summary of angles of :
multiples of
90deg
45deg
30deg

i know what the x and y values are from these.. but when i find the angles i get confused on what angle sine is .. eg with root3/2 i dont know if the angle going anti clockwise is 30 or 60 unless i use my calculator..

i hope you guys understand what im talking about... as my teacher keeps mentioning the acute angle which i have no idea on where is is ??
• Sep 21st 2007, 11:23 PM
Thomas
Acute means an angle less than 90 degrees.

The instructions say to use your unit circle diagram... do you have one?
• Sep 21st 2007, 11:56 PM
whitestrat
Yes i do have one, but just for test purposes i cant bring one in....
also i know what acute means but im just wondering where its measured from..

thanks anyway :)
• Sep 22nd 2007, 04:36 AM
topsquark
Quote:

Originally Posted by whitestrat
Yes i do have one, but just for test purposes i cant bring one in....
also i know what acute means but im just wondering where its measured from..

thanks anyway :)

All angles are measured from the positive x-axis in a counterclockwise fashion.

Reference angles are measured from either the positive or negative x-axis and are measured in either a clockwise or countclockwise fashion. You can tell if you need to go clockwise or counterclockwise by looking to see what quadrant the angle needs to fall into.

For example:
$sin(\theta) = -\frac{\sqrt{3}}{2}$

$sin(\theta)$ is the y-coordinate on the unit circle and is negative, so your angle must lie in either the third or fourth quadrants. The reference angle will be an acute angle such that
$sin(\theta) = \left | - \frac{\sqrt{3}}{2} \right | = \frac{\sqrt{3}}{2}$

So the reference angle is $60^o$. Since this needs to be in QIII or QIV we get two angles for $\theta$:
$\theta = 180^o + 60^o = 240^o$
and
$\theta = 360^o - 60^o = 300^o$

-Dan