Question

*URGENTE* Sendo, sen α = -1/3, π < α < 3π/2, calcule cos α, tg α, sec α e cos sec α.

Answer 1

Explicação passo-a-passo:

Pela relação fundamental da trigonometria:

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

$\sf \left(-\dfrac{1}{3}\right)^2+cos^2~\alpha=1$

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

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

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

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

Como esse ângulo pertence ao 3° quadrante, seu cosseno é negativo

$\sf cos~\alpha=-\sqrt{\dfrac{8}{9}}$

$\sf \red{cos~\alpha=-\dfrac{2\sqrt{2}}{3}}$

Assim:

• $\sf tg~\alpha=\dfrac{sen~\alpha}{cos~\alpha}$

$\sf tg~\alpha=\dfrac{-\frac{1}{3}}{-\frac{2\sqrt{2}}{3}}$

$\sf tg~\alpha=\dfrac{1}{3}\cdot\dfrac{3}{2\sqrt{2}}$

$\sf tg~\alpha=\dfrac{1}{2\sqrt{2}}$

$\sf tg~\alpha=\dfrac{1}{2\sqrt{2}}\cdot\dfrac{\sqrt{2}}{\sqrt{2}}$

$\sf tg~\alpha=\dfrac{\sqrt{2}}{2\cdot2}$

$\sf \red{tg~\alpha=\dfrac{\sqrt{2}}{4}}$

• $\sf \sec~\alpha=\dfrac{1}{cos~\alpha}$

$\sec~\alpha=\dfrac{1}{\frac{-2\sqrt{2}}{3}}$

$\sec~\alpha=-\dfrac{3}{2\sqrt{2}}$

$\sf sec~\alpha=-\dfrac{3}{2\sqrt{2}}\cdot\dfrac{\sqrt{2}}{\sqrt{2}}$

$\sf sec~\alpha=-\dfrac{3\sqrt{2}}{2\cdot2}$

$\sf \red{sec~\alpha=-\dfrac{3\sqrt{2}}{4}}$

• $\sf cossec~\alpha=\dfrac{1}{sen~\alpha}$

$\sf cossec~\alpha=\dfrac{1}{-\frac{1}{3}}$

$\sf \red{cossec~\alpha=-3}$