`\qquad sinα=-1/ 3; π/2<α<{3π}/2`
`=>cosα<0`
Ta có:
`\qquad sin^2α+cos^2α=1`
`=>cos^2α=1-sin^2α=1-(-1/ 3)^2=8/ 9`
$⇒\left[\begin{array}{l}cosα=\dfrac{2\sqrt{2}}{3}(loại)\\cosα=\dfrac{-2\sqrt{2}}{3}\end{array}\right.$
$\\$
`tanα={sinα}/{cosα}={-1/ 3}/{{-2\sqrt{2}}/3}=1/{2\sqrt{2}}`
`cotα=1/{tanα}=2\sqrt{2}`
Vậy: `sinα=-1/ 3;cosα={-2\sqrt{2}}/3`
`\qquad tanα=1/{2\sqrt{2}};cotα=2\sqrt{2}`
