Question

函数fx=√3(sinx)^2+sinxcosx-√3/2,设三角形三边为abc,若f(A/2+

Answer 1

解由fx=√3(sinx)^2+sinxcosx-√3/2

=√3(1-cos2x)/2+1/2sin2x-√3/2

=-√3cos2x/2+1/2sin2x

=sin(2x-π/3)

又由,若f(A/2+π/4)=1

即sin(2(A/2+π/4)-π/3)=1

即sin(A+π/2-π/3)=1

即cos(A-π/3)=1

即A-π/3=0

即A=π/3

又由a^2=b^2+c^2-2bccosA

即4=b^2+c^2-2bc×1/2

即b^2+c^2-bc=4

即(b+c)^2-3bc=4

即(b+c)^2-4=3bc≤3[(b+c)/2]^2

即(b+c)^2-4≤3/4[(b+c)]^2

即1/4[(b+c)]^2≤4

即[(b+c)]^2≤16

即b+c≤4

故b+c的最大值为4.