Figuring out the sign of sinx given cosx

[Mr Davis 97]

Given that $\cos x = -0.4927$, $\sin x = \pm \sqrt{1 - (\cos x)^2} = \pm \sqrt{1 - (-0.4927)^2} = \pm 0.8704$. How do I know which sign to choose for the correct sign?

 

[haruspex]

Given that $\cos x = -0.4927$, $\sin x = \pm \sqrt{1 - (\cos x)^2} = \pm \sqrt{1 - (-0.4927)^2} = \pm 0.8704$. How do I know which sign to choose for the correct sign?

You don't. You will need some other fact to make that choice.

 

[Mr Davis 97]

Oh. Like what extra information would I need?

 

[haruspex]

Oh. Like what extra information would I need?

That's too open-ended. What other informmation do you have? Or is this a generic question?

 

[Mr Davis 97]

Just a generic question. Would I need to know what quadrant cosx is in? Since it could be in either II or III.

 

[haruspex]

Just a generic question. Would I need to know what quadrant cosx is in? Since it could be in either II or III.

You need some way to determine whether x is in 2nd or 3rd quadrant, yes.

 

[symbolipoint]

That gives two possible answers. Refer to the Unit Circle.

 

[symbolipoint]

Given that $\cos x = -0.4927$, $\sin x = \pm \sqrt{1 - (\cos x)^2} = \pm \sqrt{1 - (-0.4927)^2} = \pm 0.8704$. How do I know which sign to choose for the correct sign?

They BOTH are correct. You make a selection according to the situation you are solving.

 

[Mark44]

Would I need to know what quadrant cosx is in?

You would need to know what quadrant x is in, not what quadrant cos(x) is in.

 

[rahul_26]

Both are correct because here sign of cosine is negative and cosine has negative values in 2nd and 3rd quadrant and sine has positive values in 2nd quadrant and negative values in 3rd quadrant. So here sign of sine depends on quadrant.