Доказателство:

[moni2003petrova]

Докажете тъждеството:
3(sin^4.α + cos^4.α) - 2(sin^6.α + cos^6.α) = 1

[KOPMOPAH]

$$3(\sin^4α + \cos^4α) - 2(\sin^6α + \cos^6α) =$$
$$=3(\sin^4α + \cos^4α)-2\underbrace{(\sin^2α + \cos^2α)}_{=1}(\sin^4α-\sin^2α \cos^2α+ \cos^4α)=$$
$$=3\sin^4α + 3\cos^4α-2\sin^4α+2\sin^2α \cos^2α-2\cos^4α=$$
$$=\sin^4α+2\sin^2α \cos^2α+\cos^4α=\underbrace{(\sin^2α + \cos^2α)^2}_{=1}=1$$

[S.B.]

Докажете тъждеството:
3(sin^4.α + cos^4.α) - 2(sin^6.α + cos^6.α) = 1

$= 3(sin^{4}\alpha + cos^{4}\alpha) - 2(sin^{6}\alpha + cos^{6}\alpha) =$
$ = 3[(sin^{2}\alpha + cos^{2}\alpha)^{2} - 2sin^{2}\alpha.cos^{2}\alpha] - 2[(sin^{2}\alpha)^{3} + (cos^{2}\alpha)^{3}] =$
[tex]= 3( 1 - 2sin^{2}\alpha.cos^{2}\alpha) - 2[(sin^{2}\alpha + cos^{2}\alpha)(sin^{4}\alpha - sin^{2}\alpha.cos^{2}\alpha + cos^{4}\alpha)] =[/tex]
[tex]= 3 - 6sin^{2}\alpha.cos^{2}\alpha - 2[(sin^{2}\alpha + cos^{2}\alpha)^{2} - 3sin^{2}\alpha.cos^{2}\alpha] =[/tex]
[tex]= 3 - 6sin^{2}\alpha.cos^{2}\alpha - 2 + 6sin^{2}\alpha.cos^{2}\alpha =[/tex]
$= 3 - 2 = $
$= 1$

[S.B.]

Съжалявам! Заедно сме писали с KOPMOPAH!
Ама май моето доказателство е малко по-различно...