Question

求这道数学题！谢谢！

求这道数学题！谢谢！以上

Answer 1

(1) f(x)=sin^2x-sin^2(x-π/6)
=1/2[2sin^2x-2sin^2(x-π/6)]
=1/2[1-cos2x+cos(2x-π/3)-1]
=1/2[cos(2x-π/3)-cos2x]
=1/2[cos2xcosπ/3+sin2xsinπ/3-cos2x]
=1/2[1/2cos2x+√3/2sin2x-cos2x]
=1/2[√3/2sin2x-1/2cos2x]
=1/2sin(2x-π/6)
π/2+2kπ=<2x-π/6<=3π/2+2kπ
π/3+kπ=<x<=5π/6+kπ
单调递减区间：[π/3+kπ,5π/6+kπ] (k∈Z)
(2) x∈[-π/3,π/4]
2x-π/6∈[-5π/6,π/3]
2x-π/6=-π/2时，f(x)取得最小值：f(x)min=1/2sin(-π/2)=-1/2
2x-π/6=π/3时，f(x)求得最大值：f(x)max=1/2sin(π/3)=√3/4

Answer 2

不会