Fórum de Matemática
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Deduza uma formula por Sen(3x) em termos de senx. Sabendo que Sen= C.O/H

olá .

sabendo que :

\(\\\\ cos(\alpha +\alpha )=cos(\alpha )*cos(\alpha)-sen(\alpha )+sen(\alpha )\\\\ cos(2\alpha )=cos^{2}(\alpha )-sen^{2}(\alpha )\)

e também que:

\(\\\\\\ sen(\alpha + \alpha)=sen(\alpha)*cos(\alpha)+sen(\alpha)*cos(\alpha) \\\\ sen(2\alpha)=2sen(\alpha)*cos(\alpha)\)

então teremos:

\(\\\\\\ sen(2\alpha+\alpha)=sen(2\alpha)*cos(\alpha)+sen(\alpha)*cos(2\alpha) \\\\\\ sen(3\alpha)=2sen(\alpha )*cos(\alpha )*cos(\alpha )+sen(a)*(cos^{2}(\alpha )-sen^{2}(\alpha ))\\\\\\ sen(3\alpha)=2sen(\alpha )*cos^{2}(\alpha )+sen(a)*cos^{2}(\alpha )-sen^{3}(\alpha )\)
\(\\\\ sen(3\alpha)=3sen(\alpha )*cos^{2}(\alpha )-sen^{3}(\alpha ) \\\\ sen(3\alpha)=3sen(\alpha )*(1-sen^{2}(\alpha ))-sen^{3}(\alpha ) \\\\ sen(3\alpha)=3sen(\alpha )-3sen^{3}(\alpha )-sen^{3}(\alpha ) \\\\ sen(3\alpha)=3sen(\alpha )-4sen^{3}(\alpha) \\\\\)

agora da relação: \(sen (\alpha )=\frac{C.O}{H}\) ficamos com:

\(\\\\ sen(3\alpha )=3(\frac{C*O}{H})-4*(\frac{C*O}{H})^{3} \\\\ sen(3\alpha )=3(\frac{C*O}{H})-4*\frac{(C*O)^{3}}{H^{3}} \\\\ sen(3\alpha )=\frac{3*C*O}{H}-\frac{4*(C*O)^{3}}{H^{3}} \\\\ sen(3\alpha )=\frac{3*C*O*H^{2}-4*C^{3}*O^{3}}{H^{3}} \\\\ sen(3\alpha )=\frac{C*O(3*H^{2}-4*C^{2}*O^{2})}{H^{3}}\)


seria isto? tens o gabarito?

att mais.