Giúp mình câu này với ạ

`A=(cosalpha+2cos2alpha+cos3alpha)/(sinalpha+2sin2alpha+sin3alpha)`

`=( (cosalpha+cos 3alpha)+2cos2alpha)/( (sinalpha+sin 3alpha)+2sin 2 alpha)`

`=(2cos2alphacosalpha+2cos2alpha)/(2sin2alphacosalpha+2sin2alpha)`

`=(2cos2alpha.(cosalpha+1))/(2sin2alpha.(cosalpha+1))`

`=(cos 2alpha)/(sin 2alpha)`

`=cot2alpha`

Đáp án:

`A = (cos alpha + 2 cos 2alpha + cos 3alpha)/(sin alpha + 2 sin 2alpha + sin 3alpha)`

   ` = ( [cos alpha + cos 3alpha] + 2 cos 2alpha)/([sin alpha + sin 3alpha] + 2sin 2alpha)`

   ` = ( 2 . cos\ (alpha + 3alpha)/2 . cos\ (3alpha - alpha)/2 + 2 cos 2alpha)/( 2 . sin\ (alpha + 3alpha)/2 . cos\ (3alpha - alpha)/2 + 2 sin2alpha)`

   ` = ( 2 . cos 2alpha . cos alpha + 2 cos 2alpha)/(2 . sin 2alpha . cos alpha + 2 sin 2alpha)`

   ` = ( 2 . cos 2alpha . (cos alpha + 1))/(2 . sin 2alpha . (cos alpha + 1))`

   ` = (cos 2alpha)/(sin 2alpha) = cot 2alpha`

Áp dụng công thức : `cos a + cos b = 2 . cos\ (a+b)/2 . cos\ (a-b)/2`

                                  `sin a + sin b = 2 . sin\ (a+b)/2 . cos\ (a-b)/2`