Question

已知函数f(x)=cos[sin(x+π÷3)-√3sin(x+π÷2)]+√3÷4 若f(θ÷2

已知函数f(x)=cos[sin(x+π÷3)-√3sin(x+π÷2)]+√3÷4
若f(θ÷2＋5π÷12)＝3÷10 (0＜θ＜π÷2 )求tanθ的值 求f(x)的最小正周期及函数g(x)＝f(－π÷2)的单调增区间

Answer 1

f(x)=cos[sin(x+π/3)-√3sin(x+π/2)]+√3/4
=cos[½sinx+√3/2·cosx-√3sin(x)]+√3/4
=cos[sin(x-π/3)]+√3/4
∵cosx是偶函数
∴cos[π+sin(x-π/3)]=cos[-sin(x-π/3)]=cos[sin(x-π/3)]
∴f(x+π)=f(x),f(x)的最小正周期是π