Olimpíadas ⇒ Trigonometria Tópico resolvido

[Babi123]

Encontre a soma:
[tex3]E=\sen x+\sen\(x+\frac{\pi}{4}\)+\sen \(x+\frac{2\pi}{4}\)+...+\sen\(x+\frac{99\pi}{4}\)[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]E=\sen x+\sen\(x+\frac{\pi}{4}\)+\sen \(x+\frac{2\pi}{4}\)+...+\sen\(x+\frac{99\pi}{4}\)[/tex3]

[Auto Excluído (ID:12031)]

[tex3]\sin(x) = \frac{e^{ix} + e^{-ix}}{2}[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]\sin(x) = \frac{e^{ix} + e^{-ix}}{2}[/tex3]

[tex3]\sum_{n=0}^{99} \sen (x + n\frac{\pi}4) =\frac12 \sum e^{ix+in\frac{\pi}{4}} + e^{-ix-in\frac{\pi}{4}} = \frac12 e^{ix} \sum_n e^{in\frac{\pi}4} + \frac12e^{-ix}\sum_n{e^{-in\frac{\pi}4}}[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]\sum_{n=0}^{99} \sen (x + n\frac{\pi}4) =\frac12 \sum e^{ix+in\frac{\pi}{4}} + e^{-ix-in\frac{\pi}{4}} = \frac12 e^{ix} \sum_n e^{in\frac{\pi}4} + \frac12e^{-ix}\sum_n{e^{-in\frac{\pi}4}}[/tex3]

da soma da PG [tex3]\sum_{n=0}^{99} e^{in\frac{\pi}4} = \frac{e^{i100*\frac{\pi}4}-1}{e^{i\frac{\pi}4-1}} = \frac{e^{25i\pi}-1}{\cos(\frac{\pi}4) + i \sen(\frac{\pi}4) -1} = -2\frac{1}{\frac{\sqrt2}2-1 + i \frac{\sqrt2}2}[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]\sum_{n=0}^{99} e^{in\frac{\pi}4} = \frac{e^{i100*\frac{\pi}4}-1}{e^{i\frac{\pi}4-1}} = \frac{e^{25i\pi}-1}{\cos(\frac{\pi}4) + i \sen(\frac{\pi}4) -1} = -2\frac{1}{\frac{\sqrt2}2-1 + i \frac{\sqrt2}2}[/tex3]

[tex3]\sum_{n=0}^{99} e^{-in\frac{\pi}4} = \frac{e^{-i100*\frac{\pi}4}-1}{e^{-i\frac{\pi}4-1}} = -2 \frac{1}{\frac{\sqrt2}2-1-i\frac{\sqrt 2}2}[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]\sum_{n=0}^{99} e^{-in\frac{\pi}4} = \frac{e^{-i100*\frac{\pi}4}-1}{e^{-i\frac{\pi}4-1}} = -2 \frac{1}{\frac{\sqrt2}2-1-i\frac{\sqrt 2}2}[/tex3]

racionalizando tudo no final: [tex3]cos(x) - (1+\sqrt2)\sen x[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]cos(x) - (1+\sqrt2)\sen x[/tex3]