Question

find cos theta if tan theta =-sqrt 3 over 3 and falls in quad 2. Will award medal if answer is given

Answer 1

Well, tangent is sine over cosine. We have the following potential fraction combinations in the 2nd quadrant: The following are given in the order of 2pi/3, 3pi/4, and 5pi/6

\[\frac{ \frac{ \sqrt{3} }{ 2 } }{ \frac{ -1 }{ 2 } }\]

\[\frac{ \frac{ \sqrt{2} }{ 2 } }{ -\frac{ \sqrt{2} }{ 2 } }\]

\[\frac{ \frac{ 1 }{ 2 } }{ \frac{ -\sqrt{3} }{ 2 } }\]

Answer 2

so the answer is?

Answer 3

Well, the 3pi/4 answer is out. Look at the division of 2pi/3 and 5pi/6 and see. Not that hard to check which one

Answer 4

the second one?

Answer 5

5pi/6 :3 So the second of the two I just mentioned, ya.

Answer 6

so negative one half?

Answer 7

i ned th answer if you want the medal

Answer 8

Its not about the medal. But the answer was 5pi/6. Since cosine is the denominator, then it would be -sqrt3/2

Answer 9

okay so how did you get that?

Answer 10

what did you divide to get that?

Answer 11

Well, there are only 3 possibilities in the 2nd quadrant. Since tangent is sine divided by cosine, I set up fractions of sin/cos for each of the 3 main points in quadrant 2. From there I see that the 2nd one I put, 3pi/4, would just give me -1. I need a sqrt3 in my answer, so it makes sense that it could be the top or bottom one. In terms of doing the division, I would do this:

\[\frac{ \frac{ 1 }{ 2 } }{ \frac{ -\sqrt{3} }{ 2 } } = \frac{ 1 }{ 2 } * \frac{ -2 }{ \sqrt{3} } = \frac{ -1 }{ \sqrt{3} } = \frac{ -\sqrt{3} }{ 3 }\]

Answer 12

At 5pi/6, the sine value is 1/2 and the cosine valueis -rt3/2. dividing those in that fashion eventually leads me to the value you're looking for.

Answer 13

Thanks man

Answer 14

Np ^_^

Answer 15

what about find cos theta if sin theta=-2/3 and falls in quadrant 4

Answer 16

@Psymon

Answer 17

I can show this as a triangle drawn in the 4th quadrant. Because sin represents the opposite side of a triangle over the hypotenuse, I can assign these two values to the appropriate sides of this triangle. I now need to solve for the missing side, which can be done through pythagorean theorem. Cosine is positive in the 4th quadrant, so there is no need to keep the negative sign with my answer, which is why I did not type it. So once you use the pythagorean theorem to find the missing side, use the fact that cosine represents the adjacent side of atriangle over the hypotenuse and this will give you the cosine valueyou need.

Answer 18

okay so whats next?

Answer 19

Do you know how to solve for the missing side?

Answer 20

not really

Answer 21

can you show me really really quick bro?

Answer 22

Well, we need pythagorean theorem.

The pythagorean theorem is a^2 + b^2 = c^2



so what I have is a missing in this case. If I want to solve for a, I can rearrange the pythagorean theorem and say that a^2 = c^2 - b^2. Therefore:

\[a = \sqrt{(3)^{2}-(2)^{2}}\]

Answer 23

so it'll be positive \[\frac{ \sqrt{5} }{ 3 }\] or \[\frac{ 2\sqrt{5} }{ 5 }\]

Answer 24

i want to say the first one @Psymon

Answer 25

Yep, the first one would be correct : )

Answer 26

okay so lets say same equation but it was in quadrant 3? would it only make it negavtive or would it be a different equation?

Answer 27

@Psymon

Answer 28

It would simply be negative.

Answer 29

ohhhh makes sense then so it would be a whole different answer?

Answer 30

Pretty much. But only because of the fact that cosine is negative in the 3rd quadrant.

Answer 31

okay thank you soo much

Answer 32

Yep yep :3

Answer 33

You're great help. I posted something about you sooo thanks and if I have anyhting I need help on. I'm coming to you bro.

Answer 34

You really are the best @Psymon