Question

设函数f(x)=根号（3）sinxcosx-cosxsin(π/2+x)-1/2

(1)求f(x)最小正周期（2）当x属于[0.π/2]时，求函数f(x)的最大值或最小值

Answer 1

y=√3sinxcosx-cosxsin（π/2+x）-1/2 =√3/2sin2x-(cosx)^2-1/2 =sin(2x-π/6)-1最小正周期：T=πf(x)的最大值：f(π/3)=sin(2*π/3-π/6)-1=0f(x)的最小值：f(0)=sin(-π/6)-1=-3/2

Answer 2

太难了

Answer 3

啥意识