12/2010. Ako je [inlmath]\cos 2\alpha=\frac{\sqrt 3}{3}[/inlmath], onda je vrednost izraza [inlmath]\sin^4\alpha+\cos^4\alpha[/inlmath] jednaka:
[inlmath]\enclose{circle}{A)}\;\frac{2}{3};\quad[/inlmath] [inlmath]B)\;\frac{3}{4};\quad[/inlmath] [inlmath]C)\;\frac{4}{5};\quad[/inlmath] [inlmath]D)\;\frac{5}{6};\quad[/inlmath] [inlmath]E)\;\frac{6}{7};\quad[/inlmath] [inlmath]N)\;\mbox{Ne znam}.[/inlmath]
[dispmath]\cos 2\alpha=\frac{\sqrt 3}{3}[/dispmath][dispmath]\sin^4\alpha+\cos^4\alpha=\left(\sin^2\alpha\right)^2+\left(\cos^2\alpha\right)^2\\
=\left(\sin^2\alpha+\cos^2\alpha\right)^2-2\sin^2\alpha\cos^2\alpha\\
=1-\sin^22\alpha\\
=\cancel{\sin^22\alpha}+\enclose{box}{\cos^22\alpha}-\cancel{\sin^22\alpha}[/dispmath][dispmath]\cos^22\alpha=\left(\frac{\sqrt 3}{3}\right)^2=\frac{3}{9}=\frac{1}{3}[/dispmath]
Gde gresim?

[dispmath]1-\frac{4}{2}\sin^2\alpha\cos^2\alpha=1-\frac{1}{2}\left(2^2\sin^2\alpha\cos^2\alpha\right)[/dispmath][dispmath]=1-\frac{1}{2}\sin^22\alpha=1-\frac{1}{2}\left(1-\cos^22\alpha\right)[/dispmath][dispmath]=1-\frac{1}{2}\left(1-\left(\frac{\sqrt 3}{3}\right)^2\right)=\frac{2}{3}[/dispmath]

Ne kapiram kako si radio, pokusacu da raspisem pa mozda shvatim, ali hvala u svakom slucaju ...

[inlmath]2[/inlmath] ispred zagrade se napise kao [inlmath]\frac{4}{2}[/inlmath] da bi moglo da bude [inlmath]2^2[/inlmath] u zagradi, a ispred zagrade ostane [inlmath]\frac{1}{2}[/inlmath].
Imas resen 1. zadatak u postu "Trigonometrijski izrazi".

ivzo je napisao:[dispmath]1-\frac{4}{2}\sin^2\alpha\cos^2\alpha=1-\frac{1}{2}\left(2^2\sin^2\alpha\cos^2\alpha\right)[/dispmath]
[dispmath]=1-\frac{1}{2}\sin^2 2\alpha=1-\frac{1}{2}\left(1-\cos^2 2\alpha\right)[/dispmath]
[dispmath]=1-\frac{1}{2}\left(1-\left(\frac{\sqrt 3}{3}\right)^2\right)=\frac{2}{3}[/dispmath]

Shvatio sam, gresku sam napravio kod pretvaranja [inlmath]2\sin^2\alpha\cos^2\alpha[/inlmath] u [inlmath]\sin^22\alpha[/inlmath].
Hvala!