Z równania [latex]sin alpha +cos alpha = frac{2}{3} [/latex] i 1 trygonometrycznej wychodzi, że: [latex]sin alpha = frac{1}{4} \ cos alpha = frac{5}{12} [/latex] Więc [latex] sin^{3} alpha +cos^{3} alpha = frac{19}{216} [/latex]
[latex]sin alpha +cos alpha =frac{2}{3}\ sin alpha =frac{2}{3}-cos alpha \ \ sin^2 alpha +cos^2 alpha =1\ (frac{2}{3}-cos alpha )^2+cos^2 alpha =1\ frac{4}{9}-frac{4}{3}cos alpha +cos^2 alpha +cos^2 alpha =1\ 2cos^2 alpha -frac{4}{3}cos alpha -frac{5}{9}=0\ Delta = (-frac{4}{3})^2-4cdot 2cdot (-frac{5}{9})=frac{16}{9}+frac{40}{9}=frac{56}{9}\ [/latex] [latex]sqrt{Delta}=sqrt{frac{56}{9}}=frac{2sqrt{14}}{3}\ cos alpha = frac{frac{4}{3}-frac{2sqrt{14}}{3}}{4}=frac{frac{4-2sqrt{14}}{3}}{4}=frac{2-sqrt{14}}{6} vee cos alpha = frac{frac{4}{3}+frac{2sqrt{14}}{3}}{4}=frac{2+sqrt{14}}{6}\ \ sin alpha =frac{2}{3}-cos alpha =frac{2}{3}-frac{2-sqrt{14}}{6}=frac{2+sqrt{14}}{6}\ vee sin alpha =frac{2}{3}-cos alpha =frac{2}{3}-frac{2+sqrt{14}}{6}=frac{2-sqrt{14}}{6}\ \ \ [/latex] [latex]sin^3 alpha +cos^3 alpha = (frac{2+sqrt{14}}{6})^3+(frac{2-sqrt{14}}{6})^3=frac{8+12sqrt{14}+84+14sqrt{14}}{216}+\ \+frac{8-12sqrt{14}+84-14sqrt{14}}{216}=frac{180}{216}=frac{5}{6}[/latex]
