Puzzle trigonométrico :: Trigonometric Puzzle

Sejam $n$ um inteiro positivo e $f(x)$ uma função trigonométrica. Determine $n$ e $f(x)$ tais que:

$\displaystyle 2f\left( \frac{\pi }{n}\right) =\sqrt{1-\frac{1}{2}\sqrt{2}}+\sqrt{1+\frac{1}{2}\sqrt{2}}$

Let $n$ be a positive integer and $f(x)$ some trigonometric function. Find $n$ and $f(x)$ such that:

$\displaystyle 2f\left( \frac{\pi }{n}\right) =\sqrt{1-\frac{1}{2}\sqrt{2}}+\sqrt{1+\frac{1}{2}\sqrt{2}}$.

Solução :: Solution

Sobre Américo Tavares

eng. electrotécnico reformado / retired electrical engineer
Esta entrada foi publicada em Matemática, Matemática-Secundário, Math, Trigonometria com as etiquetas , , , . ligação permanente.

Uma resposta a Puzzle trigonométrico :: Trigonometric Puzzle

1. Solução de Jacques Glorieux por e-mail :: Solution by Jaques Glorieux via e-mail:

$\sqrt{1-\frac{1}{2}\sqrt{2}}+\sqrt{1+\frac{1}{2}\sqrt{2}}$

$=\sqrt{\cos 0-\cos \frac{\pi }{4}}+\sqrt{\cos 0-\cos \frac{\pi }{4}}$

$=\sqrt{-2\sin \left( \frac{1}{2}\left( 0+\frac{\pi }{4}\right) \right) \sin\left( \frac{1}{2}\left( 0-\frac{\pi }{4}\right) \right) }$
$+\sqrt{2\cos \left( \frac{1}{2}\left( 0+\frac{\pi }{4}\right) \right) \cos\left( \frac{1}{2}\left( 0-\frac{\pi }{4}\right) \right) }$

$=\sqrt{2\sin ^{2}\frac{\pi }{8}}+\sqrt{2\cos ^{2}\frac{\pi }{8}}$

$=\sqrt{2}\left( \sin \frac{\pi }{8}+\cos \frac{\pi }{8}\right)$

$=2\left( \frac{\sqrt{2}}{2}\sin \frac{\pi }{8}+\frac{\sqrt{2}}{2}\cos \frac{\pi }{8}\right)$

$=2\left( \sin \frac{\pi }{4}\sin \frac{\pi }{8}+\cos \frac{\pi }{4}\cos\frac{\pi }{8}\right)$

$=2\cos \left( \frac{\pi }{4}-\frac{\pi }{8}\right)$

$=2\cos \left( \frac{\pi }{8}\right)$

$\implies f(x)=\cos x$ $\wedge$ $n=8$

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