Question

已知函数f(x)=sin(π-x)sin(π/2-x)+cos²x

(1)求函数f(x)的最小正周期
(2)当x[-π/8,3π/8]时，求函数f(x)的单调区间

Answer 1

（1）f(x)=sinxcosx+cos²x=1/2sin2x+1/2(2cos²x-1)+1/2=1/2sin2x+1/2cos2x+1/2=√2/2sin(2x+π/4)+1/2所以T=2π/2=π
(2)x[-π/8,3π/8]所以2x+π/4[0,π],所以2x+π/4[0,π/2]即增区间【-π/8，π/8】减区间【π/8,3π/8】