IME / ITA ⇒ Números Complexos Tópico resolvido

[Hanon]

Prove que:
[tex3]\sen \(\frac{2\pi}{7}\)+\sen \(\frac{4\pi}{7}\)+\sen \(\frac{8\pi}{7}\)=\frac{\sqrt{7}}{2}[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]\sen \(\frac{2\pi}{7}\)+\sen \(\frac{4\pi}{7}\)+\sen \(\frac{8\pi}{7}\)=\frac{\sqrt{7}}{2}[/tex3]

Não possuo Gabarito

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[Andre13000]

viewtopic.php?f=3&t=57497

Cópia da resolução:

Seja [tex3]\psi=\cis \frac{2\pi}{7}[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]\psi=\cis \frac{2\pi}{7}[/tex3]

[tex3]S=\sen\frac{2\pi}{7}+\sen\frac{4\pi}{7}+\sen\frac{8\pi}{7}\\
S=\frac{\psi-\psi^{-1}+\psi^2-\psi^{-2}+\psi^4-\psi^{-4}}{2i}\\
2iS=\psi+\psi^2+\psi^{4}-\psi^{-1}-\psi^{-2}-\psi^{-4}\\
\psi^7=1\\
2iS=\psi+\psi^2+\psi^4-\psi^6-\psi^5-\psi^3\\
p=\psi+\psi^2+\psi^4\\
q=\psi^6+\psi^5+\psi^3\\
1+p+q=\sum_{n=0}^6 \psi^n=\frac{\psi^7-1}{\psi-1}=0\\
p+q=-1\\
pq=\psi^4+\psi^5+\psi^6+3\psi^7+\psi^8+\psi^9+\psi^{10}\\
pq=\psi^4+\psi^5+\psi^6+3 +\psi+\psi^2+\psi^3=3+p+q=2\\
t^2+t+2=0\\
4t^2+4t+8=0\\
4t^2+4t+1=-7\\
2t+1=\pm i\sqrt{7}\\
t=\frac{-1\pm i\sqrt{7}}{2}\\
2iS=p-q>0\\
p=\frac{i\sqrt{7}-1}{2}\\
q=-\frac{i\sqrt{7}+1}{2}\\
2iS=i\sqrt{7}\\
S=\frac{\sqrt{7}}{2}\\
\sqrt{a+b}=3[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]S=\sen\frac{2\pi}{7}+\sen\frac{4\pi}{7}+\sen\frac{8\pi}{7}\\
S=\frac{\psi-\psi^{-1}+\psi^2-\psi^{-2}+\psi^4-\psi^{-4}}{2i}\\
2iS=\psi+\psi^2+\psi^{4}-\psi^{-1}-\psi^{-2}-\psi^{-4}\\
\psi^7=1\\
2iS=\psi+\psi^2+\psi^4-\psi^6-\psi^5-\psi^3\\
p=\psi+\psi^2+\psi^4\\
q=\psi^6+\psi^5+\psi^3\\
1+p+q=\sum_{n=0}^6 \psi^n=\frac{\psi^7-1}{\psi-1}=0\\
p+q=-1\\
pq=\psi^4+\psi^5+\psi^6+3\psi^7+\psi^8+\psi^9+\psi^{10}\\
pq=\psi^4+\psi^5+\psi^6+3 +\psi+\psi^2+\psi^3=3+p+q=2\\
t^2+t+2=0\\
4t^2+4t+8=0\\
4t^2+4t+1=-7\\
2t+1=\pm i\sqrt{7}\\
t=\frac{-1\pm i\sqrt{7}}{2}\\
2iS=p-q>0\\
p=\frac{i\sqrt{7}-1}{2}\\
q=-\frac{i\sqrt{7}+1}{2}\\
2iS=i\sqrt{7}\\
S=\frac{\sqrt{7}}{2}\\
\sqrt{a+b}=3[/tex3]