Question

已知函数f（x）＝根号3（sinx^2－cosx^2）－2sinxcosx⑴ 求f（x）的最小正周期...

已知函数f（x）＝根号3（sinx^2－cosx^2）－2sinxcosx⑴ 求f（x）的最小正周期 ⑵设x∈[－π/3，π/3]，求f（x）的值域和单调递增区间

Answer 1

f（x）＝根号3（sinx^2－cosx^2）－2sinxcosx
=-√3cos2x-sin2x
=-2sin(2x+π/3)
最小正周期为 2π/2=π
x∈迟启冲[－π/3，π/3]，
则 2x+π/3∈[-π/3,π]
所以码歼值域为 [-2,√3]
单调增区间：
2x+π∈[π/2,π]
x∈[-π/4,0]
所以
单调增区间为 [-π/旁樱4,0]

Answer 2

f(x)=-√3cos2x-sin2x
=-2(sin2x*cosπ/3+cos2xsinπ/3)
=-2sin(2x+π/3)
所以T=2π/2=π

-π/3<=x<=π/3
-π/3<=2x+π/3<=π/2
此范败丛纯围sin递增，郑咐则-2sin(2x+π/3)递减
所以-2sin(-π/3)=√3,-2sin(π/2)=-2
值域察咐[-2,√3]
减区间是[-π/3,π/3]