已知f(x)=cos^2x-sin^x+2sinxcosx。①求函数最小正周期②当x∈【0...
发布网友
发布时间：2024-12-01 10:55
共2个回答
热心网友
时间：2024-12-01 14:27
f（x）=cos^2x-sin^x+2sinxcosx
= cos2x+sin2x
= √2(sinπ/4cos2x+cosπ/4sin2x)
= √2sin(2x+π/4)
函数最小正周期 = 2π/2 = π
x∈【0,π/2】
2x∈【0，π】
2x+π/4∈【π/4，5π/4】
2x+π/4∈【π/4，π/2】时单调增
2x+π/4∈【π/2，5π/4】时单调减
2x+π/4=π/2时最大值f(x)max = √2
2x+π/4=5π/4时最小值f(x)min = √2*-√2/2 = -1
热心网友
时间：2024-12-01 14:25
π，√2，-1
