Question

sen beta = 4/5 e beta pertence 2ºquadrante, calcule cos(2beta).

Answer 1

Pela relação fundamental da trigonometria:

$\text{sen}^2\beta+\text{cos}^2\beta=1~\hookrightarrow~\left(\dfrac{4}{5}\right)^2+\text{cos}^2\beta=1~\hookrightarrow~\dfrac{16}{25}+\text{cos}^2\beta=1$

$\text{cos}^2\beta=1-\dfrac{16}{25}~\hookrightarrow~\text{cos}^2\beta=\dfrac{25-16}{25}~\hookrightarrow~\text{cos}^2\beta=\dfrac{9}{25}$

Como $\beta$ pertence ao $2^{\circ}$ quadrante, temos que $\text{cos}~\beta<0$ 

$\text{cos}^2\beta=\dfrac{9}{25}~\hookrightarrow~\text{cos}~\beta=-\sqrt{\dfrac{9}{25}}~\hookrightarrow~\text{cos}~\beta=-\dfrac{3}{5}$

Lembre-se que:

$\text{cos}~2\beta=\text{cos}^2\beta-\text{sen}^2\beta$

Logo:

$\text{cos}~2\beta=\left(-\dfrac{3}{5}\right)^2-\left(\dfrac{4}{5}\right)^2~\hookrightarrow~\text{cos}~2\beta=\dfrac{9}{25}-\dfrac{16}{25}~\hookrightarrow~\boxed{\text{cos}~2\beta=-\dfrac{7}{25}}$