Question

Is square root 1 minus cosine squared theta = −sin Θ true? If so, in which quadrants does angle Θ terminate?

Answer 1

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

Answer 2

That's how it actually looks

Answer 3

@phi

Answer 4

First,
when people write the square root with no sign out front, we assume it's the positive square root.

Second, there is this *very useful* identity
\[ \sin^2(x) + \cos^2(x) = 1 \]
from which we can derive

\[ \sin^2(x) = 1 - \cos^2(x) \]

Answer 5

I follow so far

Answer 6

using that identity in your
\[ \sqrt{1-\cos ^2}\theta=-\sin \theta \\
\sqrt{\sin ^2}\theta=-\sin \theta \\ \sin \theta=-\sin \theta
\]

Answer 7

How can something equal itself but negative?

Answer 8

Ohhh, So the equation is false?

Answer 9

when you see
x = -x
to solve, we add +x to both sides to get
x+x= -x + x
2x = 0
divide both sides by 2
x= 0

so the solution would be x=0

in your case
sin theta = 0

Answer 10

Excuse me, expression

Answer 11

it's an equation

Answer 12

Darn it, I was just working with an expression, and I was thinking of that. But either way, does that mean the equation is false? Because undefined is not one of the answers.

Answer 13

theta = 0 definitely works.

I am thinking about if there are other solutions...

Answer 14

I see- Would the answer choices help?

Answer 15

A. False

B. True; quadrants I & IV

C. True; quadrants II & III

D. True; quadrants III & IV

Answer 16

ok, here is how to think about it.
see https://en.wikipedia.org/wiki/Absolute_Square#Real_numbers
\[ | x | = \sqrt{x^2} \]
so we should write
\[ \sqrt{\sin ^2}\theta=-\sin \theta \\
| \sin \theta | = -\sin \theta \\ | \sin \theta | + \sin \theta=0
\]

this will be true wherever sin theta is negative.
for example, say sin(theta) = -1
then
| sin(theta) | + sin(theta)
| -1| + -1
+1 + -1
0

Answer 17

so it comes down to "what quadrants is sin(x) negative" ?

Answer 18

1 and 2?

Answer 19

Oh! negative. 3 and 4

Answer 20

yes, 3 and 4

Answer 21

So the correct answer is D?

Answer 22

this question was tricky!

Answer 23

yes, D is the answer

Answer 24

Thank you sosososo much for your help. There's no way I would have gotten the answer alone.

Answer 25

You could have tackled it this way: pick an angle in the "middle" of each quadrant (that you know the sine and cosine of, and test that angle in the equation.
That would lead to figuring out the correct quadrants.

Answer 26

Thank you! I can see how that would have worked. I actually have one more question I'm confused with. Would you possibly mind helping me out with it? I don't think it's quite as long.

Answer 27

did you post it?

Answer 28

Yes, I did. I'll mention you