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Christians4
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\[Tan^2x - \cos^2x = 1/\cos^2 - 1 - \cos^2x\]
lgbasallote
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can you rewrite that right side? is that \[\frac 1{cos^2 x - 1} - \cos ^2 x\]
Christians4
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|dw:1352207602501:dw|
lgbasallote
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you know this identity right? \[\sin^2 x = 1 - \cos^2 x\]
Christians4
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yes
lgbasallote
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wait..nevermind what i said....
Christians4
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alright
lgbasallote
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\[\frac 1{\cos^2x} - 1 - \cos^2 x\]
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?
Christians4
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yes :) I was on the right track whew :P
lgbasallote
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now combine the numerators \[\frac{1 - \cos^2 x - \cos ^ 4 x}{\cos^2 x}\]
is this what you did?
Christians4
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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?
lgbasallote
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no. that works too \[\frac{1 - \cos^2 x}{\cos^2 x} - \cos^2 x\]
yes?
Christians4
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hmm, why didnt you just totally cancel out the denominator?
Christians4
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I was left with 1-cos^2x-cos^2x :S
lgbasallote
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ah. that's different
lgbasallote
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you can't do that
HorribleAtMath
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|dw:1352208030577:dw|
Christians4
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hmm whats the reason I couldnt?
lgbasallote
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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
HorribleAtMath
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|dw:1352208100756:dw|
HorribleAtMath
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ap calc etcher tell me that
Christians4
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OH I understand
lgbasallote
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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?
Christians4
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yes! I love you :P
lgbasallote
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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?
Christians4
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should i move to the left side?
Christians4
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wo think i got it? L.S) will equel 1-cos^2/cos^2x - Cos^2x
Christians4
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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
lgbasallote
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tan^3 x = tan^2 x * tan x
lgbasallote
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that can work
lgbasallote
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or maybe tan^3 x = sin^3 x / cos^3 x
Rosh007
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\[\tan ^{2}x - \cos^{2}x =\frac{1}{ \cos^2(x)} -1- \cos^{2}x \]
\[\tan ^{2}x - \cos^{2}x =\frac{(1-\cos^{2}x) }{ \cos^2(x)} - \cos^{2}x \]
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
Christians4
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Thanks guys, lemme try this other one and i cant get it ill post a new question
Rosh007
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all the best :)