Question

Verifica egalitatea
$sin ^{6} x+cos^{6}x=1-3sin^{2}x*cos^{2}x$

Answer 1

   
$\text{Folosim formulele: } \\ a^{3}+ b^{3}=(a+b)(a^{2}-ab+ b^{2}) \\a^{2}+ b^{2} = (a+b)^{2}-2ab \\ sin^{2}x + cos^{2}x = 1$
    
$sin ^{6} x+cos^{6}x=(sin^{2} x)^{3}+(cos^{2} x)^{3}= \\ =(sin^{2} x+cos^{2}x)((sin^{2} x)^{2}-sin^{2} x*cos^{2}x+(cos^{2} x)^{2})= \\ =(1)((sin^{2} x)^{2}-sin^{2} x*cos^{2}x+(cos^{2} x)^{2})= \\ =(sin^{2} x)^{2}+(cos^{2} x)^{2}-sin^{2} x*cos^{2}x = \\ =(sin^{2} x+cos^{2} x)^{2}-2sin^{2} x*cos^{2}x-sin^{2} x*cos^{2}x= \\ =(1)^{2} -3sin^{2} x*cos^{2}x = \boxed{1 - 3sin^{2} x*cos^{2}x}$