向量OP=(2COSX+1.COS2X-SINX+1),向量OQ=(COSX,-1),f(X)=OP*OQ,求f(X)的单减区间

题目

向量OP=(2COSX+1.COS2X-SINX+1),向量OQ=(COSX,-1),f(X)=OP*OQ,求f(X)的单减区间

答案

f(x)=cosx(2cosx+1)-(cos2x-sinx+1)
=2(cosx)^2+cosx-cos2x+sinx-1
=2(cosx)^2-1-cos2x+cosx+sinx
=cos2x-cos2x+cosx+sinx
=sinx+cosx
=√2[√2sinx/2+√2cosx/2]
=√2sin(x+π/4)
2kπ+π/2≤x+π/4≤2kπ+3π/2
2kπ+π/4≤x ≤2kπ+4π/3
f(X)的单减区间[2kπ+π/4,2kπ+4π/3]