# Finding $\sin^6 x+\cos^6 x$, what am I doing wrong here?

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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Expanded $(\sin^2 x+\cos^2 x)^3$ incorrectly. Need $3\sin^4 x\cos^2 x+3\cos^4 x\sin^2 x$ as the last two terms. – André Nicolas Jun 20 '13 at 19:33
@AndréNicolas Oh yeah haha the expansion is soo wrong, I don't know how I came up with it. But if I do the expansion correctly I think I can still solve it this way. – Ovi Jun 20 '13 at 19:36
As a side note, I released this exact problem several weeks back on Brilliant. Ovi, if that is where you obtained the problem from, you can click on the link to view the solution. – Calvin Lin Jun 20 '13 at 20:06
@CalvinLin Yes thank you that is where I got it from but I wanted to do it on my own before I looked at the solution. It worked perfectly fine after I corrected my mistake. – Ovi Jun 20 '13 at 20:09
Yes, I can tell from your working above (as opposed to simply copying the question and expecting complete answers) :) – Calvin Lin Jun 20 '13 at 20:10

$(\sin^2 x + \cos^2 x)^3=\sin^6 x + \cos^6 x + 3\sin^2 x \cos^2 x$

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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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$\sin^6x + \cos^6x = (\sin^2x)^3 + (\cos^2x)^3 =(\sin^2x + \cos^2x)(\sin^4x + \cos^4x -\sin^2x\cos^2x)$

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

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

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Use $\LaTeX$ command \sin instead sin, and similar for cos. – Cortizol Jul 21 '13 at 10:32
It has to be $1 - 1 / 3$, not $1 - (1 / 3) ^ 2$, the answer is $2 / 3$. – user98213 Oct 2 '13 at 10:54
$a^3+b^3=(a+b)(a^2-ab+b^2),$ not $a^3+b^3=(a+b)(a^2+ab+b^2).$ – Cameron Buie Oct 2 '13 at 11:22