Question

已知函数f(x)=cos(2x-π/3)+sin^2 x-cos^2 x 求单调递减

Answer 1

f(x)=cos(2x-π/3)-(cos^2 x-sin^2 x)
=cos(2x-π/3)-cos2x
=2sin(2x-π/6)sinπ/6
=sin(2x-π/6)
因为y=sinx的单减区间为[π/2+2kπ,3π/2+2kπ](k为整数)
---->y=sin2x的单增区间为[π/4+kπ,3π/4+kπ](k为整数)
---->y=sin(2x-π/6)的单增区间为[π/6+kπ,2π/3+kπ]
---->y=-sin(2x-π/6)的单增区间为[π/6+kπ,2π/3+kπ]