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

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

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

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

milliex51 Group TitleBest ResponseYou've already chosen the best response.0
false? .. oh.
 2 years ago

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

experimentX Group TitleBest ResponseYou've already chosen the best response.1
1# wolfram says so 2# it is giving solution ... this is only true in certain points. not on every ponits
 2 years ago

milliex51 Group TitleBest ResponseYou've already chosen the best response.0
i don't understand..
 2 years ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
http://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
 2 years ago

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

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

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

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

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

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

experimentX Group TitleBest ResponseYou've already chosen the best response.1
\( (\tan2x + 1)  (\tan2x+1)\cos2x = \tan 2x \) \( \implies 1  (\tan2x+1)\cos2x = 0 \)
 2 years ago
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