Question

Seja, sen a = -2/5 com a € 3° quadrante, calcule sen (2a).

Answer 1

Pela relação fundamental da trigonometria:

$\text{sen}^2\alpha+\text{cos}^2\alpha=1~\hookrightarrow~\left(-\dfrac{2}{5}\right)^2+\text{cos}^2\alpha=1~\hookrightarrow~\dfrac{4}{25}+\text{cos}^2\alpha=1$

$\text{cos}^2\alpha=1-\dfrac{4}{25}~\hookrightarrow~\text{cos}^2\alpha=\dfrac{25-4}{25}~\hookrightarrow~\text{cos}^2\alpha=\dfrac{21}{25}$

Como $\alpha$ pertence ao $3^{\circ}$ quadrante, temos que $\text{cos}~\alpha<0$ 

$\text{cos}^2\alpha=\dfrac{21}{25}~\hookrightarrow~\text{cos}~\alpha=-\sqrt{\dfrac{21}{25}}~\hookrightarrow~\text{cos}~\alpha=-\dfrac{\sqrt{21}}{5}$

Lembre-se que:

$\text{sen}~2\alpha=2\cdot\text{sen}~\alpha\cdot\text{cos}~\alpha$

Logo:

$\text{sen}~2\alpha=2\cdot\left(-\dfrac{2}{5}\right)\cdot\left(-\dfrac{\sqrt{21}}{5}\right)~\hookrightarrow~\boxed{\text{sen}~2\alpha=\dfrac{4\sqrt{21}}{25}}$