Question

已知函数f(x)=cosxsin(x+π/6)-cos2x-1/4,x∈R求f(x)的单调递增区间

已知函数f(x)=cosxsin(x+π/6)-cos2x-1/4,x∈R求f(x)的单调递增区间 求f(x)在[-π/6，π/4]上的最大值和最小值

Answer 1

f(x)=cosxsin(x+π/6)-cos2x-1/4,
=cosx(√3/2sinx+1/2cosx)-cos2x-1/4,
=√3/2sinxcosx+1/2(cosx)^2-cos2x-1/4,
=√3/4sin2x+1/4(1+cos2x)-cos2x-1/4,
=√3/4sin2x-3/4cos2x
=√3/2(1/2sin2x-√3/2cos2x)
=√3/2sin(2x-π/3)
2x-π/3∈[2kπ-π/2,2kπ+π/2]单调递增
所以有:
x∈[kπ-π/12,kπ+5π/12]单调递增

因x∈[-π/6,π/4]
所以,f(x)=√3/2sin(2x-π/3) ,在2x-π/3∈[-2π/3,π/6]上
最大值=√3/4
最小值=-3/4