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2012-11-15 · 知道合伙人教育行家
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f(x)=2(sinx-cosx)cosx
=2sinxcosx-2cos²x
=sin2x-(2cos²x-1)-1
=sin2x-cos2x-1
=√2(√2/2*sin2x+√2/2*cos2x)-1
=√2sin(2x-π/4)-1
∴函数f(x)最小正周期为:T=2π/2=π
函数的单调递增区域为:-π/2+2kπ≤2x-π/4≤π/2+2kπ,k为整数
即:-π/8+kπ≤x≤3π/8+kπ,k为整数
则单调递减区间为:[-π/8+kπ,3π/8+kπ],k为整数
=2sinxcosx-2cos²x
=sin2x-(2cos²x-1)-1
=sin2x-cos2x-1
=√2(√2/2*sin2x+√2/2*cos2x)-1
=√2sin(2x-π/4)-1
∴函数f(x)最小正周期为:T=2π/2=π
函数的单调递增区域为:-π/2+2kπ≤2x-π/4≤π/2+2kπ,k为整数
即:-π/8+kπ≤x≤3π/8+kπ,k为整数
则单调递减区间为:[-π/8+kπ,3π/8+kπ],k为整数
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