Question

Se β é um ângulo pertencente ao quarto quadrante do círculo trigonométrico, de tal forma que: cos β = 3/5, então, qual é o valor de “sen β”?

Answer 1

$\Large\boxed{\begin{array}{l}\sf quando~\beta\in~4^o~quadrante,sen(\beta)<0.\\\sf cos(\beta)=\dfrac{3}{5}\implies cos^2(\beta)=\dfrac{9}{25}\\\sf \dfrac{25}{25}-\dfrac{9}{25}=\dfrac{16}{25}\\\\\sf sen^2(\beta)=\dfrac{16}{25}\\\sf sen(\beta)=-\sqrt{\dfrac{16}{25}}\\\\\sf sen(\beta)=-\dfrac{4}{5}\end{array}}$