Question

Sendo sen x = -1/3, qual das afirmações é verdadeira?
não sei nem pra que lado p fazer isso socorro

Answer 1

Explicação passo-a-passo:

Pela relação fundamental da trigonometria:

$\sf sen^2~x+cos^2~x=1$

$\sf \Big(-\dfrac{1}{3}\Big)^2+cos^2~x=1$

$\sf \dfrac{1}{9}+cos^2~x=1$

$\sf cos^2~x=1-\dfrac{1}{9}$

$\sf cos^2~x=\dfrac{9-1}{9}$

$\sf cos^2~x=\dfrac{8}{9}$

$\sf cos~x=\pm\sqrt{\dfrac{8}{9}}$

$\sf cos~x=\pm\dfrac{2\sqrt{2}}{3}$

a) Falsa

b) Falsa

$\sf sen(\pi+x)=sen~\pi\cdot cos~x+sen~x\cdot cos~\pi$

$\sf sen(\pi+x)=0\cdot\dfrac{2\sqrt{2}}{3}+\Big(-\dfrac{1}{3}\Big)\cdot(-1)$

$\sf sen(\pi+x)=0+\dfrac{1}{3}$

$\sf sen(\pi+x)=\dfrac{1}{3}$

c) Verdadeira

$\sf sen(\pi-x)=sen~\pi\cdot cos~x-sen~x\cdot cos~\pi$

$\sf sen(\pi-x)=0\cdot\dfrac{2\sqrt{2}}{3}-\Big(-\dfrac{1}{3}\Big)\cdot(-1)$

$\sf sen(\pi-x)=0-\dfrac{1}{3}$

$\sf sen(\pi-x)=-\dfrac{1}{3}$

d) Falsa

$\sf cos~\Big(\dfrac{\pi}{2}+x\Big)=cos~\Big(\dfrac{\pi}{2}\Big)\cdot cos~x-sen~\Big(\dfrac{\pi}{2}\Big)\cdot sen~x$

$\sf cos~\Big(\dfrac{\pi}{2}+x\Big)=0\cdot\dfrac{2\sqrt{2}}{3}-1\cdot\Big(-\dfrac{1}{3}\Big)$

$\sf cos~\Big(\dfrac{\pi}{2}+x\Big)=0+\dfrac{1}{3}$

$\sf cos~\Big(\dfrac{\pi}{2}+x\Big)=\dfrac{1}{3}$

Letra C