设函数f(x)=cos(2x+π/3)+sin^2x-1/2

题目

设函数f(x)=cos(2x+π/3)+sin^2x-1/2
当x属于[0,π]时,求f(x)的递减区间
当f(a-π/8)=√3/3时 求f(2a)的值

答案

f(x)=cos(2x+π/3)+sin^2x-1/2
=cos(2x+π/3)+(1-cos2x)/2-1/2
=cos2xcos(π/3)-sin2xsin(π/3)-cos2x*1/2
=-√3/2*sin2x
当x∈[0,π]时,2x∈[0,2π],f(x)的递减区间即sin2x的递增区间,显然需：
0≤2x≤π/2,或3π/2≤2x≤2π
得0≤x≤π/4,或3π/4≤x≤π
f(a-π/8)=√3/3=-√3/2*sin[2(a-π/8)]
得sin(2a-π/4)=-2/3
f(2a)=-√3/2*sin4a=-√3/2*cos(4a-π/2)=-√3/2*cos[2(2a-π/4)]
=-√3/2*[1-2sin^2 (2a-π/4)]
=-√3/2*[1-2*(-2/3)^2]
=-√3/18