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.

Who wants an easy medal!

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.

Join Brainly to access

this expert answer

SEE EXPERT ANSWER

To see the expert answer you'll need to create a free account at Brainly

cos^2(theta) -1
the question is unclear. and I give you medal for free. heheh, I don't know how to solve

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

oh thanks!
oh thanks!
Trig identities,\[\cos^2\theta-1=\sin^2\theta\]
nope
it wants me to simpify
Why not? and that is simplified
\[-\cos^2(aht^2e)\]
Oh I see my mistake.. again
yea \[-\cos^2\theta \] sorry about that :/
redo
@lala2 he gave you the formula, you replace and figure out the answer. 1= sin^2 +cos^2
what about this one sec(theta)sin(theta)cot(theta)
\[\sec \theta*\sin \theta*\cot \theta=\frac{ 1 }{ \cos \theta }*\sin \theta*\frac{ \cos \theta }{ \sin \theta }=\frac{ \sin \theta }{ \cos \theta }*\frac{ \cos \theta }{ \sin }\]
*sin(theta)

Not the answer you are looking for?

Search for more explanations.

Ask your own question