Question

Is square root 1-cos^2 ø= -sin ø if so in which quadrant does angle O terminate

Answer 1

The Trigonometric Pythagorean Theorem is always true:

$\cos^2 \theta + \sin ^2 \theta = 1$

If we solve for $\sin \theta$ we get

$\sin ^2 \theta = 1 - \cos^2 \theta$

We know

$0 \le \cos ^2 \theta \le 1$

$0 \le \sin ^2 \theta \le 1$

so both sides are clearly between zero and one.

It's actually much better to do geometry using squared sine, but that's a topic for another day. It comes from replacing angle, a complicated directed meeting of two rays, with the intersection of two lines. Among many advantages, this eliminates quadrants.

But we know sine and cosine can be positive or negative. So we need the multivalued square root:

$\sin \theta =\pm \sqrt{ 1 - \cos^2 \theta}$

The $\pm$ means that it might be plus, it might be minus, if you just tell me the cosine there's not (usually) enough information to know which.

So yes, sometimes

$\sin \theta = - \sqrt{ 1- \cos^2 \theta}$

That's a negative sine. If the cosine is positive, $\theta$ is in the fourth quadrant. If the cosine is negative, $\theta$ is in the third quadrant.