$\displaystyle sin^6x + cos^6x = 1-3sin^2x cos^2x$ how can i prove that identity...tks
$\displaystyle (\sin^2{x})^3 + (\cos^2{x})^3 =$
$\displaystyle (\sin^2{x} + \cos^2{x})(\sin^4{x} - \sin^2{x}\cos^2{x} + \cos^4{x}) =$
$\displaystyle \sin^4{x} - \sin^2{x}\cos^2{x} + \cos^4{x} =$
$\displaystyle \sin^4{x} + 2\sin^2{x}\cos^2{x} + \cos^4{x} - 3\sin^2{x}\cos^2{x} =$
$\displaystyle (\sin^2{x} + \cos^2{x})^2 - 3\sin^2{x}\cos^2{x} =$
$\displaystyle 1 - 3\sin^2{x}\cos^2{x}$
$\displaystyle \sin^4{x} + 2\sin^2{x}\cos^2{x} + \cos^4{x} - 3\sin^2{x}\cos^2{x} =$
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?
$\displaystyle \sin^4{x} + 2\sin^2{x}\cos^2{x} + \cos^4{x} - 3\sin^2{x}\cos^2{x} =$
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 $ ?