Question

已知Y=½cos²x+二分之根号三×sinxcosx+1,求值域,周期和单调区间

Answer 1

y=1/2cos^2(x)+√3/2sinxcosx+1

=1/4(1+cos2x)+√3/4sin2x+1

=√3/4sin2x+1/4cos2x+5/4

=1/2[(sin2x)(√3/2)+(cos2x)(1/2)]

=1/2sin(2x+π/6)

1)

值域：【-1/2,1/2】

2)

T=2π/2=π

3)

（i）由-π/2+2kπ≤2x+π/6≤π/2+2kπ得单调增区间：

【-π/3+kπ,π/6+kπ】

(ii)由π/2+2kπ≤2x+π/6≤3π/2+2kπ得单调减区间：

【π/6+kπ,2π/3+kπ】