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/

f(x)=cos(2x+π/3)+sin^2x-1/2
=cos2xcosπ/3-sin2xsinπ/3+(1-cos2x)/2-1/2
=1/2cos2x-√3/2*sin2x+1/2-1/2cos2x-1/2
=-√3/2sin2x
x∈[0,π]
2x∈[0,2π]
f(x)的递减区间:(0,π/2)U(3π/2,2π)
f(a-π/8)=√3/3
-√3/2sin2(a-π/8)=√3/3
-√3/2sin(2a-π/4)=√3/3
sin(2a-π/4)=-2/3
cos[2(2a-π/4)]=1-2sin²(2a-π/4)
cos(4a-π/2)=1-2sin²(2a-π/4)
cos(4a-π/2)=1-2*(-2/3)²
sin4a=1-8/9
sin4a=1/9
f(2a)=-√3/2sin4a
=-√3/2*1/9
=-√3/18