Question

Доказать
3(sin^4a+cos^4a)-2(sin^6a+cos^a)=1

Answer 1

$3(sin^4 alpha +cos^4 alpha )-2(sin^6 alpha +cos^6 alpha)=1;$
$3(sin^4alpha+cos^4alpha)-2(sin^2alpha+cos^2alpha)(sin^4alpha-sin^2alpha cos^2alpha+cos^4alpha)=$
$=3(sin^4alpha+cos^4alpha)-2(sin^4alpha-sin^2alpha cos^2alpha+cos^4alpha)=$
$=3sin^4alpha+3cos^4alpha-2sin^4alpha+2sin^2alpha cos^2alpha-2cos^4alpha=$$=sin^4alpha+2sin^2alpha cos^2alpha+cos^4alpha=(sin^2alpha+cos^2alpha)^2=1;$

Answer 2

3(sin^4a+cos^4a)-2(sin^6a + cos^6a) =  3(sin^4a+cos^4a)-2(sin^2a+cos^2a)(sin^4a-sin^2a cos^2a+ cos^4a) =  3sin^4a + 3cos^4a- 2sin^4a+2sin^2a cos^2a – 2cos^4a  = sin^4a + 2sin^2a cos^2a+cos^4a = (sin^2a+cos^2a)^2 = 1