Question

设方程12x²-πx-12π＝0的两根分别为α，β.

求cosαcosβ-√3sinαcosβ-√3cosαsinβ-sinαsinβ的值

Answer 1

设方程12x²-πx-12π＝0的两根分别为α，β.
求cosαcosβ-(√3)sinαcosβ-(√3)cosαsinβ-sinαsinβ的值

解：因为α，β是方程12x²-πx-12π＝0的根，故由维达定理，得α+β=π/12；
故cosαcosβ-(√3)sinαcosβ-(√3)cosαsinβ-sinαsinβ
=cosαcosβ-sinαsinβ-[(√3)sinαcosβ+(√3)cosαsinβ]
=cos(α+β)-(√3)sin(α+β)
=2[(1/2)cos(α+β)-(√3/2)sin(α+β)]
=2[cos(α+β)cos(π/3)-sin(α+β)sin(π/3)]
=2cos(α+β+π/3)=2cos(π/12+π/3)
=2cos(5π/12)=2cos(π/6+π/4)
=2[cos(π/6)cos(π/4)-sin(π/6)sin(π/4)]
=2[(√6-√2)/4]=(√6-√2)/2

Answer 2