Trigonometria

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[IME 2016]

    \[ \large\sum_{k=1}^{15} \textrm{cis}^{2k-1}\left( \frac{\pi}{36} \right) \]

Encontre a parte imaginária do somatório acima.

Resolução:

Primeiramente lembremos que

    \begin{align*} \textrm{cis } \theta & = \cos \theta + i \sin\theta \Rightarrow \\ \Rightarrow \textrm{cis }^{2k-1} \theta & = \cos [( 2k-1) \theta] + i \sin [( 2k-1 ) \theta]. \end{align*}

E assim

    \begin{align*} S & = \large\sum_{k=1}^{15} \textrm{cis}^{2k-1}\left( \frac{\pi}{36} \right) \\ & = \large\sum_{k=1}^{15} \textrm{cis} \left[ \frac{(2k-1)\pi}{36} \right]. \end{align*}

Se \textrm{Im} [S] é a parte imaginária de S então

    \begin{align*} \textrm{Im} [S] & = \large \sum_{k=1}^{15} \sin \left[ \frac{(2k-1)\pi}{36} \right]. \\ & = \sin \frac{\pi}{36} + \sin \frac{3\pi}{36} + \ldots +\sin \frac{29\pi}{36}.\\ \end{align*}

Multiplicando ambos os lados por -2\sin \frac{\pi}{36}

    \begin{align*} -2\sin \frac{\pi}{36} \textrm{Im} [S] & = -2\sin \frac{\pi}{36} \sin \frac{\pi}{36}-2\sin \frac{\pi}{36} \sin \frac{3\pi}{36} - \ldots\\ &-2\sin \frac{\pi}{36} \sin \frac{29\pi}{36}\\ & =-2\sin^2 \frac{\pi}{36} + \biggl[ \cos \frac{2\pi}{18}-\cos \frac{\pi}{18} \biggl]+\\ &+\biggl[ \cos \frac{3\pi}{18}-\cos \frac{2\pi}{18} \biggl] +\ldots\\ &+\biggl[ \cos \frac{15\pi}{18}-\cos \frac{14\pi}{18} \biggl] \\ & =-2\sin^2 \frac{\pi}{36} - \cos \frac{\pi}{18} + \cos \frac{15\pi}{18}\\ & =-2\sin^2 \frac{\pi}{36} - \cos \frac{2\pi}{36} + \cos \frac{5\pi}{6}\\ & =-2\sin^2 \frac{\pi}{36} - (1-2\sin^2 \frac{\pi}{36}) + \cos \frac{5\pi}{6}.\\ \end{align*}

Logo

    \begin{align*} -2\sin \frac{\pi}{36} \textrm{Im} [S] & =-1 - \frac{\sqrt3}{2}\\ \textrm{Im} [S] & = \boxed{\dfrac{2+\sqrt{3}}{4 \sin \frac{\pi}{36}}}.\\ \end{align*}

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