Ace school

with brainly

  • Get help from millions of students
  • Learn from experts with step-by-step explanations
  • Level-up by helping others

A community for students.

combine the fractions and simplify to a multiple of a power of a basic trigonometric function sinx/cot^2x-sinx/cos^2x?

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

SIGN UP FOR FREE
can i change the sinx/cos^2x into tan^2x?
Hmm no. You can take ONE of the cosines, and change it to tanx*(1/cosx) But I dont know if that's the right way to go c: hmm
\[\large \frac{ \sin x }{ \cot^2 x }-\frac{ \sin x }{ \cos^2 x }=\frac{ \sin x }{ (\frac{ \cos^2 x }{ \sin^2 x }) }-\frac{ \sin x }{ \cos^2 x }\] \[\large \frac{ \sin x }{ 1 }*\frac{ \sin^2 x }{ \cos^2 x }-\frac{ \sin x }{ \cos^2 x }\]

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

There are multiple ways to simplify it, since there are so many trig identities, so you might find a different way... but this is what I was thinking :3
\[\large \frac{ \sin^3 x - \sin x }{ \cos^2 x }\]
Confused about any of that so far? :D
how did you get sinx/1∗sin^2x/cos^2x
\[\large \frac{ \left(\frac{ a }{ b }\right) }{ \left(\frac{ c }{ d }\right) }=\frac{ a }{ b }*\frac{ d }{ c }\]
Remember how you can do that? when dividing fractions :D flip the bottom one and rewrite it as multiplication.
but the top wasn't a fraction yet it was sinx/ (cos^2x/sin^2x)
oh wait you put it over 1 to make it into a fraction huh?
\[\huge \frac{ \left(\sin x \right) }{ \left(\frac{ \cos^2 x }{ \sin^2 x } \right) }=\frac{ \left(\frac{ \sin x }{ 1 } \right) }{ \left(\frac{ \cos^2 x }{ \sin^2 x } \right) }\]
Yah hehe :3
I gotta go, so lemme post the rest of the steps, and you can at least look at them while I'm gone. I won't be here to answer questions though :c
okay
\[\large \frac{ \sin^3 x - \sin x }{ \cos^2 x }=\frac{ \sin x(\sin^2 x - 1) }{ 1-\sin^2 x }\]
Factored a sinx from each term in the top. Rewrote cos^2 as 1-sin^2 using identity :D
Now if we factor a -1 ouch of each term in the top brackets, we'll have a nice easy cancellation.
\[\large \frac{ -\sin x(1-\sin^2 x) }{ 1-\sin^2 x }=-\sin x\]
Hopefully I didn't make a mistake in there :D heh
alright well thanks

Not the answer you are looking for?

Search for more explanations.

Ask your own question