Solving Trig Questions Where cos x = 0.6

[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.

 

[neutrino]

x is given in radians.

 

[swears]

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

 

[neutrino]

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

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

 

[swears]

I converted .6 radians to 34.3774677 degree.

I don't really see the connection.

 

[TD]

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

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

<br /> \cos ^2 x + \sin ^2 x = 1 \Rightarrow \sin x = \pm \sqrt {1 - \cos ^2 x} <br />

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

 

[swears]

Thanks,

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

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

 

[neutrino]

I converted .6 radians to 34.3774677 degree.

I don't really see the connection.

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

 

[swears]

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

 

[swears]

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

, ok.

 

[TD]

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

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

 

[swears]

Thanks

 

[TD]

You're welcome