riddlingminion
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How do I factor sin(^6)x + cos(^6)x?
Solve Sin(^6)x + Cos(^6)x Factoring Problem
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lotrgreengrapes7926
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Use x^3+y^3=(x+y)(x^2-xy-y^2).
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Better
[tex]\sin^{6}x +\cos^{6}x= (\sin^{2}x +\cos^{2}x)(\sin^{4}x +\cos^{4}x)-2\sin^{2}x \cos^{2}x(\sin^{2}x +\cos^{2}x)=...=\cos^{2} 2x.[/tex]
Daniel.
[tex]\sin^{6}x +\cos^{6}x= (\sin^{2}x +\cos^{2}x)(\sin^{4}x +\cos^{4}x)-2\sin^{2}x \cos^{2}x(\sin^{2}x +\cos^{2}x)=...=\cos^{2} 2x.[/tex]
Daniel.
lotrgreengrapes7926
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Doesn't [tex]( \sin^{2} x +\cos^{2} x ) ( \sin^{4}x +\cos^{4} x ) - 2 \sin^{2}x \cos^{2} x ( \sin^{2} x +\cos^{2} x) = \sin^6 x+ \cos^6 x - \sin^2 x \cos^4 x- \sin^4 x \cos^2 x[/tex]?
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dextercioby said:
Whereby the hitherto unknown equality:
[tex]\frac{1}{8}+\frac{1}{8}=0[/tex]
is proven.
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Yes, thought it was 2 simple 2 be true. You can drop that 2, then. :d
Daniel.
Daniel.