Tính sin α .

Đáp án đúng là: D

Ta có \(\left\{ \begin{array}{l}1 + {\tan ^2}\alpha  = \frac{1}{{{{\cos }^2}\alpha }}\\\frac{{2017\pi }}{2} < \alpha  < \frac{{2019\pi }}{2}\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}1 + {\left( { - \frac{4}{3}} \right)^2} = \frac{1}{{{{\cos }^2}\alpha }}\\\frac{\pi }{2} + 504.2\pi  < \alpha  < \frac{{3\pi }}{2} + 504.2\pi \end{array} \right.\)

 →cosα=−35. Mà tanα=sinαcosα⇔−43=sinα−35→sinα=45 .