em đnag cần gấp ai giúp em với ạ…

`1)` Từ `pi/2 < x < pi => sin x > 0 ; cos x , tan x , cot x < 0`

`a) cos 2x = 3/5`

`=> 1 - 2sin^2 x = 3/5`

`=> 2sin^2 x = 2/5`

`=> sin^2 x = 1/5`

`=> sin x = (sqrt{5})/5`

Lại có : `sin^2 x + cos^2 x = 1 => 1/5 + cos^2 x = 1 => cos^2 x = 4/5 => cos x = -(2sqrt{5})/5`

`tan x = (sin x)/(cos x) = -1/2`

`b) sin(x + pi/6) = sin x . cos pi/6 + cos x . sin pi/6 = (sqrt{5})/5 . (sqrt{3})/2 + (-2sqrt{5})/5 . 1/2`

                                                                              ` = (sqrt{15} - 2sqrt{5})/10`

`cos(pi/3 - 2x) = cos pi/3 . cos 2x + sin pi/3 . sin 2x`

Lại có : `sin 2x = 2sin x . cos x = 2 . (sqrt{5})/5 . (-2sqrt{5})/5 = -4/5`

                     ` = 1/2 . 3/5 + (sqrt{3})/2 . (-4/5) = (3-4sqrt{3})/10`

`tan(2x + pi/4) = ( tan 2x + tan pi/4)/(1 - tan 2x . tan pi/4)`

Ta có : `tan 2x = (sin 2x)/(cos 2x) = -4/3`

                      ` = (-4/3 + 1)/(1 - (-4/3) . 1) = -1/7`

`c) cos x = (-2sqrt{5})/5`

Từ `pi/2 < x < pi => pi/4 < x/2 < pi/2 => cos\ x/2 , sin\ x/2 > 0`

`=> 2cos^2\ x/2 - 1 = (-2sqrt{5})/5`

`=> cos^2\ x/2 = (5 - 2sqrt{5})/10`

`=> cos\ x/2 = sqrt{ (5 - 2sqrt{5})/10}`

Từ `cos x = (-2sqrt{5})/5`

`=> 1 - 2sin^2\ x/2 = (-2sqrt{5})/5`

`=> sin^2\ x/2 = (5 + 2sqrt{5})/10`

`=> sin\ x/2 = sqrt{ (5+2sqrt{5})/10}`

Bài `2:`

`a) cos 3x - cos x + sin 2x`

` = -2sin2x . sin x + sin 2x`

` = sin 2x . (1 - 2sinx)`

`b) sqrt{2} . sin 2x + cos 5x - cos 9x`

` = sqrt{2} . sin 2x - 2 . sin(-2x) . sin 7x`

` = sqrt{2} . sin 2x + 2sin 2x . sin 7x`

` = sin 2x . (sqrt{2} + 2sin 7x)`

Bài `3:`

`a) cos 3x . cos 6x . sin 9x`

` = 1/2 . (cos 9x + cos 3x) . sin 9x`

` = 1/2 . cos 9x . sin 9x + 1/2 . cos 3x . sin 9x`

` = 1/4 . sin 18x + 1/2 . 1/2 . ( sin 12x - sin(-6x))`

` = 1/4 . sin 18x + 1/4 . (sin 12x + sin 6x)`

` = 1/4 . sin 18x + 1/4 . sin 12x + 1/4 . sin 6x`

`b) 8 cos x . sin 2x . sin 3x`

` = 4(sin 3x - sin(-x)) . sin 3x`

` = 4(sin 3x + sin x) . sin 3x`

` = 4sin^2 3x + 4sin x . sin 3x`

` = 2(1 - cos 6x) + 2(cos 2x - cos 4x)`

` = 2 - 2cos 6x + 2cos 2x - 2cos 4x`