已知函数f（x）=cos^2x+sinxcosx,x∈Rf（π/6）

网友回答

f（x）=cos^2x+sinxcosx
=1/2(1+cos2x)+1/2sin2x
=1/2sin2x+1/2cos2x+1/2
=√2/2(√2/2sin2x+√2/2cos2x)+1/2
=√2/2sin(2x+π/4)+1/2
f(x)最小正周期T=2π/2=π
f(x)max=√2/2+1/2
f(x)min=-√2/2+1/2
由2kπ-π/2≤2x+π/4≤2kπ+π/2,k∈Z
得kπ-3π/8≤x≤kπ+π/8,k∈Z
f(x)递增区间为[kπ-3π/8,kπ+π/8],k∈Z
递减区间为[kπ+π/8,kπ+5π/8],k∈Z
f(π/6)=(cosπ/6)^2+sinπ/6cosπ/6
=(√3/2)^2+1/2*√3/2
=(3+√3)/4