Question

向量m=（(√3*cosx-sinx ,1),n=(2cosx ,a-√3*),x,a∈R,a为常数 (1)求y=m*n

Answer 1

y=m*n=(√3*cosx-sinx)*(2cosx)+1*(a-√3)

=2√3*cos^(2)x-2sinxcosx+a-√3

=√3(2cos^(2)x-1)-2sinxcosx+a

=√3cos2x-sin2x+a

=2cos[2x+(π/6)]+a

X∈(0 ,π/2)时,2x+(π/6) ∈(π/6 ,7π/6),

则cos[2x+(π/6)]∈[-1,1/2],

所以f(x)的最小值为-2+a=-2,∴a=0.