已知函数f(x)=2(sinx-cosx)cosx+1,x∈R. (2)求函数f(x)在区间[π/8,3π/4]上的单调区间和最大值与最小值。
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f(x)=2(sinx-cosx)cosx+1
=2sinxcosx-2cos²x+1
=sin2x-cos2x
=√2sin(2x-π/4)
x∈[π/3, 3π/4]
2x-π/4∈[5π/12, 5π/4]
所以2x-π/4∈[5π/12, π/2] 即x∈[π/3, 3π/8]时，单调递增
2x-π/4∈[π/2, 5π/4] 即x∈[3π/8, 3π/4]时，单调递减
f(x)最大=f(3π/8)=√2
f(x)最小=f(3π/4)=√2*(-√2/2)=-1
=2sinxcosx-2cos²x+1
=sin2x-cos2x
=√2sin(2x-π/4)
x∈[π/3, 3π/4]
2x-π/4∈[5π/12, 5π/4]
所以2x-π/4∈[5π/12, π/2] 即x∈[π/3, 3π/8]时，单调递增
2x-π/4∈[π/2, 5π/4] 即x∈[3π/8, 3π/4]时，单调递减
f(x)最大=f(3π/8)=√2
f(x)最小=f(3π/4)=√2*(-√2/2)=-1
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