A community for students.
Here's the question you clicked on:
 0 viewing

This Question is Closed

milliex51
 2 years ago
Best ResponseYou've already chosen the best response.0do I solve it like this: \[(\sin ^{2}x)(1+\tan ^{2x}) \] \[\sin ^{2x}\div1\times(sinx \div cosx)\] \[\sin ^{2x}cosx \div sinx\]

milliex51
 2 years ago
Best ResponseYou've already chosen the best response.0x is suppose to be down, oops.

FoolForMath
 2 years ago
Best ResponseYou've already chosen the best response.0Are you sure that is true: http://www.wolframalpha.com/input/?i=+%28%281cos+2x%29%281%2Btan+2x%29%29+%3D%3D+tan+2x

experimentX
 2 years ago
Best ResponseYou've already chosen the best response.1http://www.wolframalpha.com/input/?i=is+%281cos2x%29%281%2Btan2x%29+equal+to+tan2x

milliex51
 2 years ago
Best ResponseYou've already chosen the best response.0how would I know when it's false on a test?

experimentX
 2 years ago
Best ResponseYou've already chosen the best response.11# wolfram says so 2# it is giving solution ... this is only true in certain points. not on every ponits

experimentX
 2 years ago
Best ResponseYou've already chosen the best response.1http://www.wolframalpha.com/input/?i=+%28%281cos+2%28pi%2F3%29%29%281%2Btan+2%28pi%2F3%29%29%29+%3D%3D+tan+2%28pi%2F3%29 it is not true for x=pi/3

experimentX
 2 years ago
Best ResponseYou've already chosen the best response.1sin(x) = cos(pi/2  x) < put any value of x, you will always find it true!!

FoolForMath
 2 years ago
Best ResponseYou've already chosen the best response.0I don't know how this TrueQ works: http://www.wolframalpha.com/input/?i=TrueQ%5Bsin+%5E2+x+%2B+cos+%5E2+x+%3D+1%5D

experimentX
 2 years ago
Best ResponseYou've already chosen the best response.1LOL .. i found it a little while ago

TransendentialPI
 2 years ago
Best ResponseYou've already chosen the best response.0Why not just expand the left side and do some cancelling?

milliex51
 2 years ago
Best ResponseYou've already chosen the best response.0that's what I tried but then there's going to be no sinx at the bottom?

experimentX
 2 years ago
Best ResponseYou've already chosen the best response.1It is equivalent to .. proving \( \cos 2x ( 1 +\tan2x)  1 = 0\) \( \implies \cos2x + \sin2x + 1 = 0\) < which is a false identity!!

experimentX
 2 years ago
Best ResponseYou've already chosen the best response.1\( (\tan2x + 1)  (\tan2x+1)\cos2x = \tan 2x \) \( \implies 1  (\tan2x+1)\cos2x = 0 \)
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.