Question

sendo sen x=3/5 (x do 2° quadrante), calcule o sen 2x, cos 2x e tg 2x

Answer 1

Explicação passo-a-passo:

${sen}^{2} x + {cos}^{2} x = 1 \\ {cos}^{2} x = 1 - (\frac{3}{5})^{2} \\ {cos}^{2} x =1 - \frac{9}{25} \\ {cos}^{2} x = \frac{25 - 9}{25} \\ {cos}^{2} x = \frac{16}{5} \\ {cos}x = \sqrt{ \frac{16}{25} } \\ {cos}x = \frac{4}{5}$

sen 2x = 2 sen X. cos X =

$= 2. \frac{3}{5} . \frac{4}{5} = \frac{24}{25}$

8

$cos \: 2x = {cos}^{2} x - {sen}^{2} x \\ = \frac{16}{25} - \frac{9}{25} = \\ \frac{7}{25}$

tg x = sen x / cos x

tg x = 3/5 / 4/5

tg x = 3/4

tg 2x = 2.tg x / 1-tg x²

tg 2x = 2.3/4 / 1 - (3/4)²

tg 2x = 3/2 / 1 - 9/16

tg 2x = 3/2 / 7/16

tg 2x = 3/2 . 16/7

tg 2x = 48/14

tg 2x = 24/7