Question

Sabendo que pin/2<0>pin e sen 0=2.raiz de 2/3,calcule:
a)cos 0           b)cossec 0           c)cotg 0

Answer 1

$sen \theta=\frac{2\sqrt2}{3} \\ \\ cos^2 \theta=1-sen^2 \theta \\ \\ cos^2 \theta = 1 - \frac{8}{9}=\frac{1}{9} \\ \\ cos \theta=\frac{1}{3} \\ \\ Mas \ \frac{\pi}{2}< \theta < \pi \\ \\ \boxed{cos \theta = -\frac{1}{3}}$

$\boxed{cossec \ \theta=\frac{1}{sen \ \theta}=\frac{1}{\frac{2\sqrt2}{3}}=\frac{3}{2\sqrt2}=\frac{3\sqrt2}{4}} \\ \\ \\ \boxed{cotg \ \theta=\frac{cos \ \theta}{sen \ \theta}=\frac{-\frac{1}{3}}{\frac{2\sqrt2}{3}}=-\frac{1}{2\sqrt2}=-\frac{\sqrt2}{4}}$