# basic trigonometric identities

• March 6th 2006, 03:19 AM
Liyo
basic trigonometric identities
I don't know where to begin to do stuff like this-

if sec theta=3/2, find cos theta.

or
use the given information to determine the exact trig value

cos theta= -2/3, 90 degress<theta<180 degress; sin theta
:confused:
• March 6th 2006, 04:51 AM
topsquark
Quote:

Originally Posted by Liyo
I don't know where to begin to do stuff like this-

if sec theta=3/2, find cos theta.

or
use the given information to determine the exact trig value

cos theta= -2/3, 90 degress<theta<180 degress; sin theta
:confused:

The first one isn't so much a trig identity as it is a definition:
$cos\theta=\frac{1}{sec\theta}=\frac{1}{3/2}=2/3$

For the second one, we use probably the most standard trig identity:
$sin^2\theta + cos^2\theta = 1$
So if we have the cosine...
$sin^2\theta+(-2/3)^2=1$
$sin^2\theta+4/9=1$
$sin^2\theta=5/9$
$sin\theta=\pm \sqrt{5/9}$

Now, we are looking for an angle between 90 and 180 degrees. The sine function is positive in this interval. So $sin\theta=\sqrt{5/9}$. Specifically, we will be finding a value for theta between 0 and 90 degrees, so we are considering the angle we will find to be a reference angle.

I don't know of any "nice" angle such that the sine of theta will be this value. So find theta using a calculator. Now, this is the reference angle measured from 180 degrees, so we subtract this angle from 180. I get about 131.8 degrees.

-Dan
• March 10th 2006, 05:57 AM
c_323_h
if you're having trouble with it, just draw a picture and break the problem into smaller parts. you know that $cos$ is between 90 and 180 degrees. and that $cos \theta =-2/3$. so draw a picture. find the length of the other leg using the pythagorean theorem and find sin.