ITA - Trigonometria

por Márcia_Queiroz_ Dom 23 Fev 2014, 11:15

A expressão 2{sen[x + (11/2)π] + cotg²x} tg (x/2) é equivalente a

a) [cos x - sen²x] cotg x
b) [sen x - cos x] tg x
c) [cos x² - senx] cotg²x
d) [1 - cotg²x] sen x
e) [1 - cotg²x] sen x

por PedroCunha Dom 23 Fev 2014, 13:04

Olá.

A expressão correta é (e a resolução):

$\frac{2 \cdot \left[ \sin \left( x + \frac{11\pi}{2} \right) + cot^2x \right] \cdot \tan \left( \frac{x}{2} \right)}{ 1 + \tan^2 \left( \frac{x}{2} \right)} \\\\\circ \sin \frac{11\pi}{2} = \sin \frac{3\pi}{2} = -1 \\\\\circ \cos \frac{11\pi}{2} = \cos \frac{3\pi}{2} = 0 \\\\ \circ \sin \left( x + \frac{11\pi}{2} \right) = -\cos x \\\\\circ 1 + \tan^2 \left(\frac{x}{2} \right) = sec^2 \left( \frac{x}{2} \right) = \frac{1}{\cos^2 \left( \frac{x}{2} \right) } \\\\\circ \tan \left( \frac{x}{2} \right) = \frac{\sin \left( \frac{x}{2} \right)}{\cos \left( \frac{x}{2} \right)} \\\\ \circ \cot^2x = \frac{\cos^2x}{\sin^2x} \\\\ \star \frac{2 \cdot \left[ \sin \left( x + \frac{11\pi}{2} \right) + cot^2x \right] \cdot \tan \left( \frac{x}{2} \right)}{ 1 + \tan^2 \left( \frac{x}{2} \right)} = \frac{2 \cdot (\frac{\cos^2x}{\sin^2x} - \cos x) \cdot \frac{\sin \left(\frac{x}{2} \right)}{\cos \left(\frac{x}{2} \right)} }{\frac{1}{\cos^2x \left( \frac{x}{2} \right)}} \therefore \\\\ 2 \cdot \cos \left( \frac{x}{2} \right) \cdot \sin \left( \frac{x}{2} \right) \cdot \left[ \frac{\cos^2x - \cos x \cdot \sin^2x}{\sin^2x} \right] \therefore \\\\ \sin x \cdot \left( \frac{\cos x \cdot (\cos x - \sin^2x)}{\sin^2x}\right) \therefore \frac{\cos x}{\sin x} \cdot (\cos x - \sin^2x) \Leftrightarrow ctg x \cdot (\cos x -\sin^2x)$

Att.,
Pedro