Question

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

Answer 1

Yes, the equation √(1 - cos^2 ø) = -sin ø is true, and angle ø terminates in the first and second quadrants.

Step-by-step explanation:

Yes, the equation √(1 - cos^2 ø) = -sin ø is true.

To prove this, we can start by squaring both sides of the equation:

(1 - cos^2 ø) = sin^2 ø

Next, we can use the trigonometric identity sin^2 ø + cos^2 ø = 1 to simplify the equation:

(1 - cos^2 ø) = 1 - cos^2 ø

Since both sides of the equation are equal, the original equation √(1 - cos^2 ø) = -sin ø is true for all values of ø.

As for the quadrant in which angle ø terminates, we can determine that by solving the equation -sin ø = 0. This equation is true when ø = 0 or ø = 180 degrees. Therefore, angle ø terminates in the first and second quadrants.

Answer 2

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 √( 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 = - √( 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.