Cho sin α = $\frac{1}{3},$ với 90° < α < 180°. Tính cos2α + 3tan2α.

sin²α = ${\left(\frac{1}{3}\right)}^2=\frac{1}{9}$

Ta có: sin²α + cos²α = 1

Suy ra: cos²α = 1 – sin²α = $1-\frac{1}{9}=\frac{8}{9}$

cos α = $\pm \sqrt{\frac{8}{9}}=\pm \frac{2\sqrt{2}}{3}$

Vì 90° < α < 180° nên cos α = $-\frac{2\sqrt{2}}{3}$

cos2α = 2cos²α – 1 = $\frac{7}{9}$

sin2α = 2.sinα.cosα = 2.$2.\frac{1}{3}.-\frac{2\sqrt{2}}{3}=\frac{-4\sqrt{2}}{9}$

tan2α = $\frac{sin2\alpha }{\cos 2\alpha }=\frac{-4\sqrt{2}}{9}:\frac{7}{9}=\frac{-4\sqrt{2}}{7}$

Vậy cos2α + 3tan2α =  $\frac{7}{9}$+$3.\frac{-4\sqrt{2}}{7}=\frac{49-108\sqrt{2}}{63}$