函数y=1/2sin(2x + pai/3)-sinxcosx的单调递减区间?

函数y=1/2sin(2x + pai/3)-sinxcosx的单调递减区间?

y=1/2sin(2x+π/3)-sinxcosx=1/2(1/2sin2x+√3/2cos2x)-1/2sin2x=1/2(-1/2sin2x+√3/2cos2x)=1/2(cosπ/6cos2x-sinπ/6sin2x)=1/2cos(2x+π/6)当2x+π/6∈[0+2kπ,π+2kπ](k∈Z)时,y单调递减即0+2kπ≤2x+π/6≤π+2kπ-π/12+kπ≤x≤5π/12+kπy的单调递减区间是[-π/12+kπ,5π/12+kπ](k∈Z)======以下答案可供参考======供参考答案1:y=1/2sin(2x + π/3)-sinxcosx=1/2sin(2x + π/3)-1/2sin2x=√3/4cos2x-1/4sin2x=1/2sin(π/3-2x)2kπ+π/22kπ+π/6kπ+π/12kπ-π/12>x>kπ-7π/12

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