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Christians4Best ResponseYou've already chosen the best response.0
\[Tan^2x  \cos^2x = 1/\cos^2  1  \cos^2x\]
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.1
can you rewrite that right side? is that \[\frac 1{cos^2 x  1}  \cos ^2 x\]
 one year ago

Christians4Best ResponseYou've already chosen the best response.0
dw:1352207602501:dw
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.1
you know this identity right? \[\sin^2 x = 1  \cos^2 x\]
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.1
wait..nevermind what i said....
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.1
\[\frac 1{\cos^2x}  1  \cos^2 x\] convert to common denominators first \[\implies \frac 1{\cos^2 x}  \frac{\cos^2 x}{\cos^2 x}  \frac{\cos^4 x}{\cos^2 x}\] agree?
 one year ago

Christians4Best ResponseYou've already chosen the best response.0
yes :) I was on the right track whew :P
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.1
now combine the numerators \[\frac{1  \cos^2 x  \cos ^ 4 x}{\cos^2 x}\] is this what you did?
 one year ago

Christians4Best ResponseYou've already chosen the best response.0
yes but Instead if making it cos^4x I cancelled out the top bracket Cos^2x with the bottom one...which is probably where I went wrong?
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.1
no. that works too \[\frac{1  \cos^2 x}{\cos^2 x}  \cos^2 x\] yes?
 one year ago

Christians4Best ResponseYou've already chosen the best response.0
hmm, why didnt you just totally cancel out the denominator?
 one year ago

Christians4Best ResponseYou've already chosen the best response.0
I was left with 1cos^2xcos^2x :S
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.1
ah. that's different
 one year ago

HorribleAtMathBest ResponseYou've already chosen the best response.0
dw:1352208030577:dw
 one year ago

Christians4Best ResponseYou've already chosen the best response.0
hmm whats the reason I couldnt?
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.1
observe what i did earlier \[\frac 1{\cos^2 x}  1  \cos^2 x \implies \frac 1{\cos^2 x}  \frac{\cos^2 x}{\cos^2 x}  \frac{\cos^4 x}{\cos^2 x}\] each term has its own denominator
 one year ago

HorribleAtMathBest ResponseYou've already chosen the best response.0
dw:1352208100756:dw
 one year ago

HorribleAtMathBest ResponseYou've already chosen the best response.0
ap calc etcher tell me that
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.1
if you cancel out the cos^2 x below cos^4 x, you can only do it there, and not in the other terms. make sense?
 one year ago

Christians4Best ResponseYou've already chosen the best response.0
yes! I love you :P
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.1
so anyway, the RHS simplifies to \[\frac{1  \cos^2 x}{\cos^2 x}  \cos^2 x\] i assume you know how to proceed form tehre?
 one year ago

Christians4Best ResponseYou've already chosen the best response.0
should i move to the left side?
 one year ago

Christians4Best ResponseYou've already chosen the best response.0
wo think i got it? L.S) will equel 1cos^2/cos^2x  Cos^2x
 one year ago

Christians4Best ResponseYou've already chosen the best response.0
Ls = Rs Sweet! Thanks for your help! I also found a few more I am having trouble with...this next question has a tan^3x which I haven't used in a while i forget what i can do with it
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.1
tan^3 x = tan^2 x * tan x
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.1
or maybe tan^3 x = sin^3 x / cos^3 x
 one year ago

Rosh007Best ResponseYou've already chosen the best response.0
\[\tan ^{2}x  \cos^{2}x =\frac{1}{ \cos^2(x)} 1 \cos^{2}x \] \[\tan ^{2}x  \cos^{2}x =\frac{(1\cos^{2}x) }{ \cos^2(x)}  \cos^{2}x \] which is \\[\tan ^{2}x =\frac{(1\cos^{2}x) }{ \cos^2(x)} \] which is \[\tan ^{2}x =\frac{(\sin^{2}x) }{ \cos^2(x)}\] which satisfis the relation
 one year ago

Christians4Best ResponseYou've already chosen the best response.0
Thanks guys, lemme try this other one and i cant get it ill post a new question
 one year ago
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