Cứu t bay ơiiiiiiiii

Ta có: `sin x =+-\sqrt{1-cos^2x}`

`=+-\sqrt{1-(3/5)^2}=+-4/5`

Mà `x in (-pi/2; 0)` nên `sin x <0`

`=> sin x=-4/5`

Áp dụng công thức nhân ba, ta có:

`@ cos 3x=4cos^3x-3cosx`

`= 4.(3/5)^3-3. 3/5`

`= 108/125-9/5=-117/125`

`@ sin 3x =3sin x -4sin^3x`

`= 3.(-4/5)-4.(-4/5)^3`

`= -12/5+256/125=-44/125`

Ta có: `sin^2x+cos^2x=1`

`=>sin^2x=1-(3/5)^2=16/25`

`=>sinx=+- 4/5`

Vì `x in (-pi/2; 0)`

`=>sinx<0`

`=>sinx=-4/5`

Khi đó: `cos 3x=cos(2x+x)=cos 2xcosx-sin 2x sin x`

`=(2cos^2 x-1). cosx-2sin^2x cosx`

`=[2. (3/5)^2-1]. 3/5-2. (-4/5)^2. 3/5`

`=-117/125`

`sin 3x=sin(2x+x)=sin2xcos x+cos 2x sin x`

`=2sinxcos^2x+(2cos^2 x-1). sinx`

`=2. (-4/5). (3/5)^2+[2. (3/5)^2-1]. (-4/5)`

`=-44/125`