Alguém ajuda-me.Não sei mesmo.... Pergunta de ideia dejaruaba20

jaruaba20 @jaruaba20

March 2022 1 92 Report

Alguém ajuda-me.

Não sei mesmo

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Niiya

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Veja que o denominador sempre dobra, logo essa soma trata-se de uma soma de termos de uma progressão geométrica

Dando atenção à P.G:

$P.G~\left(\frac{\pi}{3},\frac{\pi}{6},\frac{\pi}{12},...\right)\\\\\\a_{1}=\dfrac{\pi}{3}~~~e~~~a_{2}=\dfrac{\pi}{6}$

Calculando a razão da P.G:

$q=\dfrac{a_{2}}{a_{1}}=\dfrac{\left(\frac{\pi}{6}\right)}{\left(\frac{\pi}{3}\right)}=\dfrac{\pi}{6}\cdot\dfrac{3}{\pi}=\dfrac{3}{6}~~~\therefore~~~\boxed{\boxed{q=\dfrac{1}{2}}}$
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A soma dos termos dessa P.G é uma soma infinita, pois a razão da P.G está entre 0 e 1

A soma será dada pela fórmula:

$S_{n}=\dfrac{a_{1}}{1-q}\\\\\\S_{n}=\dfrac{\left(\frac{\pi}{3}\right)}{1-\frac{1}{2}}\\\\\\S_{n}=\dfrac{\left(\frac{\pi}{3}\right)}{\left(\frac{1}{2}\right)}\\\\\\S_{n}=\dfrac{\pi}{3}\cdot\dfrac{2}{1}\\\\\\\boxed{\boxed{S_{n}=\dfrac{2\pi}{3}}}$

Logo:

$cos\left(\frac{\pi}{3}+\frac{\pi}{6}+\frac{\pi}{12}+...\right)=cos~(S_{n})\\\\\\cos\left(\frac{\pi}{3}+\frac{\pi}{6}+\frac{\pi}{12}+...\right)=cos\left(\frac{2\pi}{3}\right)\\\\\\cos\left(\frac{\pi}{3}+\frac{\pi}{6}+\frac{\pi}{12}+...\right)=cos\left(\frac{2\cdot180\º}{3}\right)\\\\\\\cos~\left(\frac{\pi}{3}+\frac{\pi}{6}+\frac{\pi}{12}+...\right)=cos~120\º$

Como cos 120º = - cos 60º:

$cos~\left(\frac{\pi}{3}+\frac{\pi}{6}+\frac{\pi}{12}+...\right)=-cos~60\º\\\\\\\boxed{\boxed{cos~\left(\frac{\pi}{3}+\frac{\pi}{6}+\frac{\pi}{12}+...\right)=-\dfrac{1}{2}}}$

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