已知函数f(x)=sinxcosx-√3cos平方x+2╱√3在x∈[六分之兀,6分之七兀时的值
问题描述：
已知函数f(x)=sinxcosx-√3cos平方x+2╱√3在x∈[六分之兀,6分之七兀时的值
答
f(x)=sinxcosx-√3cos²x+√3/2
=1/2*sin2x-√3(cos2x+1)/2+√3/2
=1/2*sin2x-√3/2*cos2x
=sin2xcosπ/3-cos2xsinπ/3
=sin(2x-π/3)
x∈[π/6,7π/6]
2x∈[π/3,7π/3]
2x-π/3∈[0,2π]
f(x)=sin(2x-π/3)
sin(2x-π/3)∈[-1,1]
所以f(x)=sinxcosx-√3cos²x+√3/2在x∈[π/6,7π/6]时的值为：[-1,1]