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forma trigonométrica, operações com complexos https://forumdematematica.org/viewtopic.php?f=20&t=6426 |
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Autor: | PedroCunha [ 04 jul 2014, 16:30 ] |
Título da Pergunta: | Re: forma trigonométrica, operações com complexos |
Olá,nsm. Façamos alguns cálculos auxiliares: \(\begin{cases} z = -8+6i \therefore z^2 = 64 - 96i + 36i^2 \therefore z^2 = 28 - 96i \\ \overline{z} = -8-6i \\ w = \frac{-i \cdot (28-96i)}{-8-6i} \therefore w = \frac{-28i + 96i^2}{-8-6i} \therefore w = \frac{-28i - 96}{-8-6i} \therefore \\\\ w = \frac{-14i - 48}{-4-3i} \therefore w = \frac{(-14i - 48) \cdot (-4+3i)}{(-4-3i) \cdot (-4+3i)} \therefore w = \frac{56i - 42i^2 + 192 - 144i}{4^2 - 3^2i^2} \therefore w = \frac{234-88i}{25} \\\\ \rho_z = \sqrt{64 + 36} = 10 \\ \cos \alpha = \frac{-8}{10} = -\frac{4}{5} \\\\ \sin \alpha = \frac{6}{10} = \frac{3}{5} \end{cases}\) Agora, basta testar as alternativas: *Mais alguns cálculos auxiliares: \(\begin{cases} \cos\left( 3\alpha - \frac{\pi}{2} \right) = \cos 3\alpha \cdot \cos \frac{\pi}{2} + \sin 3\alpha \cdot \sin \frac{\pi}{2} = \sin 3\alpha \\\\ \sin \left( 3\alpha - \frac{\pi}{2} \right) = \sin 3\alpha \cdot \cos \frac{\pi}{2} - \sin \frac{\pi}{2} \cdot \cos 3\alpha = -\cos 3\alpha \\\\ \cos \left( \alpha - \frac{\pi}{2} \right) = \sin \alpha \\\\ \sin \left( \alpha - \frac{\pi}{2} \right) = -\cos \alpha \\\\ \cos (3\alpha) = 4\cos^3\alpha - 3\cos \alpha \therefore \cos(3\alpha) = 4 \cdot -\frac{64}{125} - 3 \cdot -\frac{4}{5} \therefore \cos (3\alpha) = \frac{44}{125} \\\\ \sin (3\alpha) = 3\sin x - 4\sin^3x \therefore \sin(3\alpha) = 3 \cdot \frac{3}{5} - 4 \cdot \frac{27}{125} \therefore \sin(3\alpha) = \frac{117}{125} \end{cases}\) Alternativa a: \(10cis\left( 3\alpha - \frac{\pi}{2} \right) = 10 \cdot \left( \sin 3 \alpha - i \cdot \cos 3\alpha \right) \therefore 10 \cdot \left( \frac{117}{125} - i \cdot \frac{44}{125} \right) \therefore 2 \cdot \left( \frac{117}{25} - i \cdot \frac{44}{25} \right) \\\\ \Leftrightarrow \frac{234}{25} - i \cdot \frac{88}{25}\) Matamos o exercício, :D. Qualquer dúvida é só falar. Att., Pedro |
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