Question

高中数学题，第（17）题，详解谢谢！

Answer 1

解1f(x)=sin(x-3π/2)cos(π/2-x)+cosxcos(π-x)
=-sin(3π/2-x)sin(x)-cosxcos(x)
=cosxsinx-cosxcosx
=1/2sin2x-(1+cos2x)/2
=1/2sin2x-1/2cos2x-1/2
=√2/2(√2/2sin2x-√2/2cos2x)-1/2
=√2/2sin(2x-π/4)-1/2
故T=2π/2=π
(2)由x属于[π/4,3π/4]
则2x属于[π/2,3π/2]
则2x-π/4属于[π/4,5π/4]
则sin(2x-π/4)属于[-√2/2,√2/2]
则√2/2sin(2x-π/4)属于[-1/2,1/2]
则√2/2sin(2x-π/4)-1/2属于[-1,0]
则f(x)属于[-1,0]
故函数的值域我[-1,0].