Question

函数y=sinx(sinx+cosx)的最小正周期 值域 单调递增区间？

Answer 1

y=sinx(sinx+cosx)
=sin²x+sinxcosx
=(1-cos2x)/2+1/2sin2x
=1/2[sin2x-cos2x]+1/2
=√2/2[sin2xcos45°-cos2xsin45°]+1/2
=√2/2sin(2x-45°)+1/2
∴最小正周期为T=2π/2=π；值域为【-√2/2+1/2,√2/2+1/2】
单调递增区间为 ：2kπ-π/2,3,y=sin²x+sinxcosx=(1-cos2x)/2+sin2x/2=1/2(sin2x-cos2x)+1/2=√2/2sin(2x-π/4)+1/2
T=2π/2=π
值域为[ (1-根号2)/2,(1+根号2)/2 ]
单调递增区间为-π/2+2π ≤ 2x-π/4 ≤ π/2+2π
自己算一下啊X∈( kπ-π/8 , ...,1,最小正周期是2π,0,