# Solving Trig Functions?

• Nov 9th 2008, 06:57 AM
cjh824
Solving Trig Functions?
I have a question where I have to solve for the cos, tan, csc, sec, and cot when given a value of sin(theta). One of them is like this:

sin(theta) = 1/3 0 < theta < pi/2

Not a clue where to go with this.
Any help would be appreciated.
• Nov 9th 2008, 07:10 AM
masters
Quote:

Originally Posted by cjh824
I have a question where I have to solve for the cos, tan, csc, sec, and cot when given a value of sin(theta). One of them is like this:

sin(theta) = 1/3 0 < theta < pi/2

Not a clue where to go with this.
Any help would be appreciated.

$\displaystyle \sin \theta = \frac{1}{3} \ \ \ \ 0 < \theta < \frac{\pi}{2}$

You are looking for ratios in the 1st quadrant.

$\displaystyle \sin \theta=\frac{y}{r} \ \ \ \ \csc\theta=\frac{r}{y}$

$\displaystyle \cos \theta=\frac{x}{r} \ \ \ \ \sec\theta=\frac{r}{x}$

$\displaystyle \tan\theta=\frac{y}{x} \ \ \ \ \cot\theta=\frac{x}{y}$

You already know r and y. Use $\displaystyle r^2=x^2+y^2$ to find x.

Then, you can find all the other ratios.
• Nov 9th 2008, 07:11 AM
cjh824
Kinda looks familiar to what we were doing in class. Does that mean r^2 - y^2 = x^2?

Thanks a million too by the way.
• Nov 9th 2008, 07:16 AM
masters
Quote:

Originally Posted by cjh824
Kinda looks familiar to what we were doing in class. Does that mean r^2 - y^2 = x^2?

Thanks a million too by the way.

That would be correct!(Rock)