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Câu `1.` 

Xét `alpha in pi < alpha < (3 pi)/2`

`-> alpha in` góc phần tư thứ III

`-> Sin alpha, Cos alpha <0`

Ta có : `Sin^2 alpha + Cos^2 alpha = 1`

`=> Cos^2 alpha = 1 - Sin^2 alpha `

`=> Cos alpha = sqrt{1 - Sin^2 alpha}`

`=> Cos alpha = sqrt{1 - (-3/5)^2}`

`=> Cos alpha = sqrt{1 - 9/25}`

`=> Cos alpha = -4/5 (` Vì `Cos alpha <0)`

`* Sin alpha = -3/5` (đề bài)

`** \ cos 2alpha=2cos^2 alpha-1=2. (-4/5)^2-1=7/25`

`** \ tan 2alpha=(sin 2alpha)/(cos 2alpha)=(2 sin alpha cos alpha)/(cos 2 alpha)=(2. (-3/5). (-4/5))/(7/25)=24/7`

Câu `2` 

Cho `Cos alpha = 1/5`

`** cos 2alpha=2cos^2 alpha-1=2. (1/5)^2-1=-23/25`

`** cos 4alpha=2cos^2 2alpha-1=2. (-23/25)^2-1=(433)/(625)`

`1)`

Ta có: `sin^2 alpha+cos^2 alpha=1`

`=>cos^2 alpha=1- (-3/5)^2=16/25`

`=>cosalpha=+- 4/5`

Vì `pi<alpha< (3pi)/2=> cos alpha<0`

`=>cosalpha= -4/5`

`@\ cos 2alpha=2cos^2 alpha-1=2. (-4/5)^2-1=7/25`

`@\ tan 2alpha=(sin 2alpha)/(cos 2alpha)=(2 sin alpha cos alpha)/(cos 2 alpha)=(2. (-3/5). (-4/5))/(7/25)=24/7`

`2)`

Ta có: `cos alpha=1/5`

`=>cos 2alpha=2cos^2 alpha-1=2. (1/5)^2-1=-23/25`

`=>cos 4alpha=2cos^2 2alpha-1=2. (-23/25)^2-1=(433)/(625)`