Question

a) Se tg x =5/4, calcule sem x, cos x e sec x
b) Se sec x =2raiz quadrada de 3/3, calcule sen x, cos x, tg x e cossec x

Answer 1

a) vamos considerar que a tgx pertença ao 1ºquadrante.

${ \sec }^{2}x = 1 + { \tan }^{2} x \\ { \sec }^{2}x = 1 + {( \frac{5}{4} )}^{2} = 1 + \frac{25}{16} \\ { \sec}^{2}x = \frac{41}{16}$

$\sec(x) = \sqrt{ \frac{41}{16} } = \frac{ \sqrt{41} }{ \sqrt{16} } = \frac{ \sqrt{41} }{4}$

$\cos(x) = \frac{1}{ \sec(x) } \\ = \frac{1}{ \frac{ \sqrt{41} }{4} } = \frac{4}{ \sqrt{41} } = \frac{4 \sqrt{41} }{41}$

$\sin(x) = \cos(x). \tan(x) \\ \sin(x) = \frac{4 \sqrt{41} }{41}. \frac{5}{4} = \frac{5 \sqrt{41} }{41}$

b)

$\cos(x) = \frac{1}{ \sec(x) } \\ = \frac{3}{2 \sqrt{3} } = \frac{3 \sqrt{3} }{2.3} = \frac{ \sqrt{3} }{2}$

${ \tan }^{2}x = { \sec }^{2}x - 1 \\ = {( \frac{2 \sqrt{3} }{3} )}^{2} - 1 = \frac{4}{3} - 1 = \frac{1}{3}$

$\tan(x) = \frac{ \sqrt{3} }{ 3}$

$\csc(x) = \frac{1}{sen(x)} = 2$

$\sin(x) = \cos(x). \tan(x) \\ \sin(x) = \frac{ \sqrt{3} }{2}. \frac{ \sqrt{3} }{3} = \frac{1}{2}$