$\alpha$ thuộc góc phần tư thứ tư => sin$\alpha$ <0

$sin^{2}$ $\alpha$ =1-$cos^{2}$ $\alpha$ =1-($\frac{5}{13}$ )²

<=> sin$\alpha$ =-12/13

• sin2$\alpha$ =2sin$\alpha$ cos$\alpha$ =2×($\frac{-12}{13}$ )×$\frac{5}{13}$ =$\frac{-120}{169}$ 

• cos2$\alpha$ =2$cos^{2}$ $\alpha$ -1=2($\frac{5}{13}$ )²-1=$\frac{-119}{169}$ 

• tan2$\alpha$ =$\frac{sin2a}{cos2a}$= $\frac{120}{119}$ 

`+) sin alpha = sqrt(1-cos^2) = -12/13 ( (3pi)/2 < alpha < 2pi)`

`=> tanalpha = -12/5`

`+) sin2 alpha = 2sinalphacosx = - 120/169`

`+) cos 2alpha = 1-2sin^2 alpha = -119/169`

`+) tan 2 alpha = (2tanalpha)/(1-tan^2alpha) = 120/119`