# Simple Trig Question

1. Jun 5, 2006

### swears

If cos x = 0.6, and 3Pie/2 < x < 2Pie, then find

a. sin (x) and B. sin (-x)

I figured out x to be 5.357 by trial and error but Is there another way to do this?

Is there some sort of rule that can help me out?

I tried using Cos(x)^2 + Sin(x)^2 = 1, but I ended up getting a different answer.

2. Jun 5, 2006

3. Jun 5, 2006

### swears

I'm sorry, but could you be more specific. I'm not sure what that means

Last edited: Jun 5, 2006
4. Jun 5, 2006

### neutrino

What you found was in degrees, i.e cos(53.13º) ~ 0.6. But the x in the problem is given in terms of radian - http://mathworld.wolfram.com/Radian.html

5. Jun 5, 2006

### swears

I converted .6 radians to 34.3774677 degree.

I don't really see the connection.

6. Jun 5, 2006

### TD

This is a good idea, since you are given cos(x) and wish to find sin(x).

$$\cos ^2 x + \sin ^2 x = 1 \Rightarrow \sin x = \pm \sqrt {1 - \cos ^2 x}$$

And: sin(-x) = -sin(x).

7. Jun 5, 2006

### swears

Thanks,

I get .8, but I think the answer is -0.8 for a.

How do I know to use the - instead of +?

8. Jun 5, 2006

### neutrino

Actually I'm wondering why you found the angle in the first place. Just do what TD said.

9. Jun 5, 2006

### swears

Oh, nevermind. I think I see it. It's because of the 3Pie/2 < x < 2Pie right?

10. Jun 5, 2006

### swears

:rofl: , ok.

11. Jun 5, 2006

### TD

Indeed, sin(x) is negative in the fourth quadrant.

12. Jun 5, 2006

### swears

Thanks

13. Jun 5, 2006

### TD

You're welcome