Tính s i n α

B

Ta có:\[\tan \alpha .\cot \alpha = 1 \Rightarrow \cot \alpha = \frac{1}{{\frac{3}{4}}} = \frac{4}{3}\]

Ta lại có:\[\frac{1}{{{{\sin }^2}\alpha }} = 1 + {\cot ^2}\alpha = 1 + {\left( {\frac{4}{3}} \right)^2} = \frac{{25}}{9} \Rightarrow \sin \alpha = \pm \frac{3}{5}\].

Do\[ - \pi < \alpha < - \frac{\pi }{2}\]nên \[\sin \alpha < 0\].  Vậy\[\sin \alpha = - \frac{3}{5}\].