题目

已知函数f(x)=cos(2x-π/3)+sin^2 x-cos^2 x
1.求单减区间 2.若f（a）=3/5,2a是第一象限角,求sin2a

提问时间：2020-11-09

答案

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]
f(a)=3/5---->sin(2a-π/6)=3/5
---->cos(2a-π/6)=4/5
sin(2a-π/6+π/6)=sin(2a-π/6)cosπ/6+cos(2a-π/6)sinπ/6
=3/5 * √3/2 + 4/5 * 1/2
=(1/10)*(3√3+4)
即sin2a=(1/10)*(3√3+4)