Question

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

Answer 1

@dunnitagainn

Answer 2

Okay so this is going to be a trigonometric identity, have you covered those yet?

Answer 3

Sorry, doesn't ring a bell. :/

Answer 4

so these can be considered the laws, and they are also good to memorize, there are a few though

Answer 5

just give me one second

Answer 6

Okay

Answer 7

is the cosine^2 theta under the square root with the 1?

Answer 8

yeah like this:

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

Answer 9

okay this one is a little tricky I'm just trying to find the right way to explain it.

Answer 10

Take your time :)

Answer 11

okay, so negative sine is the reciprocal of sine meaning cosecant. CSC theta is just 1 over sine theta. have you covered these yet? CSC, SEC, COT?

Answer 12

I've seen them very little...

So, what your saying is?
\[\cos^{-1} \theta =\frac{ 1 }{ \sin \theta }\]

Answer 13

no, -cos theta and cos^-1 theta are different. the first is the reciprocal and the second is the inverse and they would perform different functions.

Answer 14

cos^-1 is actually equal to Arccos which is what you get on your calculator when you press 2nd then cosine.

Answer 15

Ahh, okay. I see now.

Answer 16

so sin^-1 is different than csc as well, because the first would be arcsin (the inverse of sin)

Answer 17

in the problem did it say -sin, or sin^-1?

Answer 18

it sad -sin

Answer 19

okay, so
sqrt(1-cos^2theta) = -sin theta
we know because of a pythagorean identity that sin^2theta+cos^2theta=1, we can rewrite that to say cos^2theta= 1-sin^2theta because we want to use the same parts here, so we want all sin
so now we have:
sqrt(1-1-sin^2theta)= -(sin) theta
we are going to square both sides to get rid of the square root,
now we have
1-1-sin^2theta = -sin^2theta
the ones cancel out so we are left with
-sin^2theta = -sin^theta so we know they are equal

Answer 20

that last part is supposed to be -sin^2theta

Answer 21

and then since we know that sin is negative and sin= y, and y is negative in the third and fourth quadrants, we know that it must terminate in one of the two. and since earlier cosine was positive, i think we can assume the fourth, i am unsure about that part though

Answer 22

here is a link that should be really helpful http://tutorial.math.lamar.edu/pdf/Trig_Cheat_Sheet.pdf

Answer 23

did that all make sense??