Oblicz wartość wyrażenia sin^6alfa + cos^6alfa + 3sin^2alfa*cos^2alfa wiedząc, że alfa jest kątem ostrym.

Wzór:  (a+b)^3 = a^3 + 3ab^2 + 3a^2b + b^3  [latex]sin^6alpha+cos^6alpha+3sin^2alphacos^2alpa = \ \ = (sin^2alpha+cos^2 alpha)^3-3sin^2 alpha cos ^4alpha-3sin ^4alpha cos^2 alpha+ \ +3sin^2alpha cos^2 alpha= \ \ = 1^3-3sin^2alphacos^2alpha(cos^2alpha+sin^2alpha-1)= \ \ = 1-3 sin^2 alpha cos^2 alpha cdot (1-1)= \ \ = 1 - 3 sin^2 alpha cos^2alpha cdot 0 = \ \ = 1-0=1[/latex]