Open study

is now brainly

With Brainly you can:

  • Get homework help from millions of students and moderators
  • Learn how to solve problems with step-by-step explanations
  • Share your knowledge and earn points by helping other students
  • Learn anywhere, anytime with the Brainly app!

A community for students.

Find cos(theta) if tan(theta) = (-2/3) and sin(theta) > 0

Mathematics
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

Find \[\cos \theta if \tan \theta = -2/3 and \sin \theta >0\]
Here, you know that the tangent is opp/adjacent. Since the sine is positive, it must be in either the first or second quadrant right?
Yes.

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

So the opposite leg must be positive, therefore since the tangent is negative the adjacent leg must be the negative one. So what quadrant is that?
Oh so you know o and a, you just need to find the h, and you can solve it :P
Yep.
Lets see what ya got when ya get it.
Thats 2nd quad
Okay:)
Yep.
\[\frac{ -3 }{ \sqrt{13}}\]
Thats it.
:D thank you!
No prob.
\[\frac{ x \sqrt{\left| x ^{2}-4 \right|} }{ x ^{2}-4 }-1>0\]

Not the answer you are looking for?

Search for more explanations.

Ask your own question