Question

докажите,что верно равенство:sin^2x+sin^4x+cos^2x-cos^4x=1-cos2x

Answer 1

$sin^2x+sin^4x+cos^2x-cos^4x=1-cos2x$
$cos^2x-cos^4x+1-cos^2x+sin^4x=1-cos2x$
$cos^2x-cos^4x+1-cos^2x+(1-cos^2x)^2=1-cos2x$
$cos^2x-cos^4x+1-cos^2x+1-2cos^2x+cos^4x=1-cos2x$
$2-2cos^2x=1-cos2x$
$2-2cos^2x=1-2cos^2x-1$
$2-2cos^2x=2-2cos^2x$

                                                   ч.т.д