$\displaystyle sin^6x + cos^6x = 1-3sin^2x cos^2x$ how can i prove that identity...tks

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- Nov 24th 2008, 06:03 PMjuanfe_zodiacprove trigonometric identities...
$\displaystyle sin^6x + cos^6x = 1-3sin^2x cos^2x$ how can i prove that identity...tks

- Nov 25th 2008, 07:35 AMskeeter
$\displaystyle (\sin^2{x})^3 + (\cos^2{x})^3 =$

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

$\displaystyle \sin^4{x} - \sin^2{x}\cos^2{x} + \cos^4{x} =$

$\displaystyle \sin^4{x} + 2\sin^2{x}\cos^2{x} + \cos^4{x} - 3\sin^2{x}\cos^2{x} =$

$\displaystyle (\sin^2{x} + \cos^2{x})^2 - 3\sin^2{x}\cos^2{x} =$

$\displaystyle 1 - 3\sin^2{x}\cos^2{x}$ - Nov 25th 2008, 05:44 PMjuanfe_zodiac
$\displaystyle \sin^4{x} + 2\sin^2{x}\cos^2{x} + \cos^4{x} - 3\sin^2{x}\cos^2{x} =$

can u explain me what did you do in that step? i didnt get that one...

why didnt you use $\displaystyle \sin^4{x} + 1\sin^2{x}\cos^2{x} + \cos^4{x} - 2\sin^2{x}\cos^2{x} =$

i know that we want that 3 but can we add terms just like that? - Nov 25th 2008, 06:13 PMskeeter
$\displaystyle \sin^4{x} + 2\sin^2{x}\cos^2{x} + \cos^4{x} - 3\sin^2{x}\cos^2{x} =$

**can u explain me what did you do in that step? i didnt get that one...**

I needed the middle term $\displaystyle + 2\sin^2{x}\cos^2{x}$ so that I could factor the $\displaystyle \sin^4{x} + \cos^4{x}$ ... to compensate adding it, I also had to also subtract it, resulting in the $\displaystyle -3\sin^2{x}\cos^2{x}$ term.

**why didnt you use $\displaystyle \sin^4{x} + 1\sin^2{x}\cos^2{x} + \cos^4{x} - 2\sin^2{x}\cos^2{x} =$**

why do that? it won't factor.

**i know that we want that 3 but can we add terms just like that?**

sure can ... isn't $\displaystyle a^2 - ab + b^2 = a^2 + 2ab + b^2 - 3ab $ ? - Nov 25th 2008, 06:15 PMjuanfe_zodiac