I have $\sin 2x=\frac 23$ , and I"m supposed to 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 $\sin x=\frac{1}{3\cos x}$ and $\cos x=\frac{1}{3\sin x}$ .

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

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 do wrong? Thanks.

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


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


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

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

or $1-3\sin^2x\cos^2x = 1-3\left(\dfrac13\right)^2 = \dfrac23$.


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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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$\begingroup$ It has to be $1 - 1 / 3$, not $1 - (1 / 3) ^ 2$, the answer is $2 / 3$. $\endgroup$
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