Olimpíadas ⇒ (Baltic Way) - Trigonometria Tópico resolvido

[theblackmamba]

Prove que:

[tex3]sin^3 18^{\circ} + sin^2 18^{\circ} = \frac{1}{8}[/tex3]

[victoria]

Olá theblackmamba!

Vamos lá:

[tex3]8sen^{3}18+ 8sen^{2}18=1[/tex3]

[tex3]8sen^{2}18( sen18+1)=1[/tex3]

[tex3]1+ sen18= \frac{1}{8} . \frac{1}{sen^{2}18}[/tex3]

[tex3]2sen54.cos36= \frac{1}{8} . \frac{1}{sen^{2}18}[/tex3]

[tex3]2sen(90-36).cos36= \frac{1}{8} . \frac{1}{sen^{2}18}[/tex3]

[tex3]2cos^{2}36= \frac{1}{8} . \frac{1}{sen^{2}18}[/tex3]

[tex3]16cos^{2}36.sen^{2}18=1[/tex3]

[tex3]cos36.sen18=\frac{1}{4}[/tex3]

[tex3]\frac{sen36}{2cos18}.cos36=\frac{1}{4}[/tex3]

[tex3]2cos18=4sen36.cos36[/tex3]

[tex3]2cos18=2sen72[/tex3]

[tex3]cos18=sen(90-18)[/tex3]

[tex3]cos18=cos18[/tex3]

C.Q.D.

Abraço.Espero ter ajudado..