作业帮问数学题呗?
来源：学生作业帮助网 编辑：作业帮 时间：2024/05/11 04:28:03
问数学题呗?
问数学题呗?
问数学题呗?
y=(1+cos2x)/2+(sin2x)/2
=(sin2x+cos2x)/2+1/2
=√2/2*sin(2x+π/4)+1/2
所以值域是[(-√2+1)/2,(√2+1)/2]
解
y=sinxcosx+cos²x
=1/2(2sinxcosx)+1/2(2cos²x-1)+1/2
=1/2sin2x+1/2cos2x+1/2
=1/2×√2(√2/2sin2x+√2/2cos2x)+1/2
=√2/2(sin2xcosπ/4+cos2xsinπ/4)+1/2
=√2/2sin(2x+π/4)+1/2
∵sin(2x+π/4)∈[-1.1]
∴y∈[1/2-√2/2,1/2+√2/2]
即是其值域
y=(cos2x+1)/2+sin2x/2
=√2/2sin(2x+π/4)+1/2
-1≤sin(2x+π/4)≤1
所以y是值域为[(1-√2)/2,(1+√2)/2]
y=（cosx)^2+sinxcosx
=1/2(1+cos2x)+1/2sin2x
=1/2(sin2x+cox2x)+1/2
=sqrt2/2sin(2x+pai/4)+1/2
因为-1=