Finding $Sin^6 X+Cos^6 X$, What Am I Doing Wrong Here?

     
I have $sin 2x=frac 23$ , & I"m supposed khổng lồ express $sin^6 x+cos^6 x$ as $frac ab$ where $a, b$ are co-prime positive integers. This is what I did:

First, notice that $(sin x +cos x)^2=sin^2 x+cos^2 x+sin 2x=1+ frac 23=frac53$ .

Now, from what was given we have $sin x=frac13cos x$ & $cos x=frac13sin x$ .

Next, $(sin^2 x+cos^2 x)^3=1=sin^6 x+cos^6 x+3sin^2 x cos x+3cos^2 x sin x$ .

Now we substitute what we found above from the given:

$sin^6 x+cos^6+sin x +cos x=1$

$sin^6 x+cos^6=1-(sin x +cos x)$

$sin^6 x+cos^6=1-sqrt frac 53$

Not only is this not positive, but this is not even a rational number. What did I vị wrong? Thanks.

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asked Jun 20, 2013 at 19:31
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OviOvi
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$(sin^2 x + cos^2 x)^3=sin^6 x + cos^6 x + 3sin^2 x cos^2 x$


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answered Jun 20, 2013 at 19:36
user67803user67803
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Should be $(sin^2 x+cos^2 x)^3=1=sin^6 x+cos^6 x+3sin^4 x cos^2 x+3cos^4 x sin^2 x$


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answered Jun 20, 2013 at 19:35
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MaazulMaazul
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$sin^6x + cos^6x = (sin^2x)^3 + (cos^2x)^3 =(sin^2x + cos^2x)(sin^4x + cos^4x -sin^2xcos^2x)$

$sin^4x+cos^4x -sin^2xcos^2x = (sin^2x + cos^2x)^2 - 2sin^2xcos^2x -sin^2xcos^2x$

or $1-3sin^2xcos^2x = 1-3left(dfrac13 ight)^2 = dfrac23$.


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edited Oct 2, 2013 at 17:43
answered Jul 21, 2013 at 8:31
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ShobhitShobhit
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$egingroup$ It has khổng lồ be $1 - 1 / 3$, not $1 - (1 / 3) ^ 2$, the answer is $2 / 3$. $endgroup$
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