anonymous
  • anonymous
prove the identity: [\left( \cos2x \over \sin x \right) = \left( \cot^2x-1 \over \csc x \right)\]
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
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.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
\[\left( \cos2x \over \sin x \right) = \left( \cot^2x-1 \over \csc x \right)\]
Mertsj
  • Mertsj
|dw:1325730453048:dw|
anonymous
  • anonymous
i got it until (cos^2 over sinx) - (sinx) after your second step i multiplied deno and numerator by sinx. but how did you get cos^2-sin^2x over sinx and how is it same as cos^2x over sinx?

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

Mertsj
  • Mertsj
Because cos(2x) = cos^2(x)-sin^2(x)
Mertsj
  • Mertsj
Look at your trig double angle formulas
anonymous
  • anonymous
ohh aha! thanks. where did sin^2x come from
Mertsj
  • Mertsj
|dw:1325731291201:dw|
anonymous
  • anonymous
yea i meant shouln't you get cos^2-sinx over sinx instead of cos^2x -sin^2x over sinx
Mertsj
  • Mertsj
And...|dw:1325731384713:dw|
anonymous
  • anonymous
i am so sorry.. shouldn't you be dividing them by sinx not sin^2x? hahah :P
Mertsj
  • Mertsj
|dw:1325731527945:dw|
anonymous
  • anonymous
thanks! i always thought |dw:1325731485027:dw|
Mertsj
  • Mertsj
|dw:1325731638760:dw|
anonymous
  • anonymous
hehe i learned something 8p im always so confused .. : Sdfksj skJ anyway okay keep going please
Mertsj
  • Mertsj
|dw:1325731751062:dw|
anonymous
  • anonymous
ohhhhhhh!!!!!!!!!!!!!!! you can always work out right side instead of the left side too right? thanks!!!!!!!!!!!!!!! :P
anonymous
  • anonymous
Thanks alot for being patient with me :) and your explanation is great and quite fast! :)
Mertsj
  • Mertsj
You can work on either side when proving an identity or both sides. You just can't work across the equal sign like adding the same thing to both sides or something like that.
Mertsj
  • Mertsj
Well I could go faster if I knew how to make fractions with the equation thing.
anonymous
  • anonymous
you just type out 'over'
Mertsj
  • Mertsj
ty
anonymous
  • anonymous
uw. p.s. are you a tutor? like to help students? anywho when you see identity questions how do you know what to do and which side to solve first?
Mertsj
  • Mertsj
I used to teach math in high school for 32 years and now I am retired and tutor students who come to my house. I can usually solve identities.
anonymous
  • anonymous
oh wow, i see. well thanks again :) your help is always appreciated! :)
Mertsj
  • Mertsj
yw
anonymous
  • anonymous
i got stuck again.. \[\left( cosx \over 1+sinx \right) = \left( 1-sinx \over cosx \right)\] what should be the nxt step ?

Looking for something else?

Not the answer you are looking for? Search for more explanations.