o valor de f

por nanda22 Sáb Nov 09 2013, 13:40

O valor de $f(\frac{\pi }{6})$ vale quanto
Pessoal, tem algo que eu estou errando, nao sei onde, olhem:

$f(x)= \frac{\frac{1}{tgx}-1}{\sqrt{3}+2senx}$

$f(\frac{\pi }{6})= \frac{\frac{1}{\frac{tg\pi }{6}}-1}{\frac{\sqrt{3}+2sen\frac{\pi }6}}$
$\frac{\frac{1}{\frac{\sqrt{3}}{3}}-1}{\sqrt{3}+2\frac{1}{2}}$

$\frac{= \frac{1}{\frac{\sqrt{3}}3{}}-\frac{\frac{\sqrt{3}}{3}}{\frac{\sqrt{3}}{3}}}{\sqrt{3}}$

$\frac{\frac{1}{\frac{\sqrt{3}}{3}}-\frac{\frac{\sqrt{3}}{3}}{\frac{\sqrt{3}}{3}}}{\frac{\frac{}{}}{\sqrt{3}}}$

$\frac{1-\frac{\sqrt{3}}{3}}{\sqrt{3}}$

$1-\frac{\sqrt{3}}{3}. \frac{1}{\sqrt{3}}=\frac{1-\sqrt{3}}{3\sqrt{3}}$

Pessoal, nao sei como terminar !!!

A resposta dá: 2- raiz de 3

por PedroCunha Sáb Nov 09 2013, 14:31

Veja:

$\\f(x) = \frac{\frac{1}{tgx} - 1}{\sqrt3 + 2senx} \rightarrow \text{Para } x= \frac{\pi}{6}:\\\\f\left(\frac{\pi}{6} \right ) = \frac{\frac{1}{\frac{\sqrt3}{3}} - 1}{\sqrt3 + 2 \cdot \frac{1}{2}} \therefore f\left(\frac{\pi}{6} \right ) = \frac{\frac{3}{\sqrt3} - 1}{\sqrt3 + 1} \therefore \\\\ f\left(\frac{\pi}{6} \right ) = \frac{\frac{3 - \sqrt3}{\sqrt3}}{\sqrt3 + 1} \therefore f\left(\frac{\pi}{6} \right ) = \frac{\left(\frac{3-\sqrt3}{\sqrt3} \right ) \cdot (\sqrt3 - 1)}{(\sqrt3 + 1) \cdot (\sqrt3 -1)} \therefore \\\\ f\left(\frac{\pi}{6} \right ) = \frac{\frac{3\sqrt3 - 3 - \sqrt{3^2} + \sqrt3}{\sqrt3}}{2} \therefore f\left(\frac{\pi}{6} \right ) = \frac{4\sqrt3 - 6}{2\sqrt3} \therefore \\\\ f\left(\frac{\pi}{6} \right ) = \frac{(4\sqrt3 - 6) \cdot 2\sqrt3}{2\sqrt3 \cdot 2\sqrt3} \therefore f\left(\frac{\pi}{6}\right) = \frac{24 - 12\sqrt3}{12} \therefore \\\\ f\left(\frac{\pi}{6} \right ) = 2 - \sqrt3$