Question

A、B为三角形ABC的两个内角,向量a=2cos(A+B)/2向量i+sin(A-B)/2向量j(其中向量i,j是互相垂

Answer 1

由已知,a^2=[2cos(A+B)/2]^2+[sin(A-B)/2]^2

=2[1+cos(A+B)]+(1/2)[1-cos(A-B)]

=5/2+2(cosAcosB-sinAsinB)-(1/2)(cosAcosB+sinAsinB)=3/2,

∴（3/2）cosAcosB-(5/2)sinAsinB=-1,

请检查题目