求函数f(x)=cox(x π/4)xsinx的单调区间与最值?
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f(x)=cos(x+π/4)·sinx?
=[cosx·cos(π/4)-sinxcos(π/4)·sinx
=(√2/2)[sinxcosx-sin²x]
=(√2/4)[sin2x+cos2x-1]
=(√2/4)·√2sin(2x+π/4)-√2/4
=½sin(2x+π/4)-√2/4
∴最大值=½-√2/4=(2-√2)/4
最小值=-½-√2/4=-(2+√2)/4
sinx的单调递增区间x∈(2kπ-π/2,2kπ+π/2)
sinx的单调递减区间x∈(2kπ+π/2,2kπ+3π/2)
将2x+π/4看成整体：
f(x)单调递增区间2x+π/4∈(2kπ-π/2,2kπ+π/2)→x∈(kπ-3π/8,2kπ+π/8)
f(x)单调递减区间2x+π/4∈(2kπ+π/2,2kπ+3π/2)→x∈(kπ+π/8,2kπ+5π/8)
=[cosx·cos(π/4)-sinxcos(π/4)·sinx
=(√2/2)[sinxcosx-sin²x]
=(√2/4)[sin2x+cos2x-1]
=(√2/4)·√2sin(2x+π/4)-√2/4
=½sin(2x+π/4)-√2/4
∴最大值=½-√2/4=(2-√2)/4
最小值=-½-√2/4=-(2+√2)/4
sinx的单调递增区间x∈(2kπ-π/2,2kπ+π/2)
sinx的单调递减区间x∈(2kπ+π/2,2kπ+3π/2)
将2x+π/4看成整体：
f(x)单调递增区间2x+π/4∈(2kπ-π/2,2kπ+π/2)→x∈(kπ-3π/8,2kπ+π/8)
f(x)单调递减区间2x+π/4∈(2kπ+π/2,2kπ+3π/2)→x∈(kπ+π/8,2kπ+5π/8)
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