# Prove that sin^6x + cos^6x = 1

I have \$sin 2x=frac 23\$ , & I"m supposed to lớn 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\$ và \$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 vì chưng wrong? Thanks.

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algebra-precalculus trigonometry
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asked Jun 20, 2013 at 19:31

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

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

ShobhitShobhit
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\$egingroup\$ It has to lớn be \$1 - 1 / 3\$, not \$1 - (1 / 3) ^ 2\$, the answer is \$2 / 3\$. \$endgroup\$
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Oct 2, 2013 at 10:54

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