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

f(x)=sin(π-x)sin(π/2-x)+cos²x
f(x)=sinxcosx+cos²x
=1/2sin2x+1/2+1/2cos2x
=1/2(sin2x+cos2x)+1/2
=√2/2sin(2x+π/4)+1/2
T=2π/2=π
答：函数f(x)的最小正周期是π 画图出来可知 x∈ [-π/8,π/8]单调递增 [π/8,3π/8]单调递减=1/2(sin2x+cos2x)+1/2 =√2/2sin(2x+π/4)+1/2怎么得到的？你看这个的变换就知道了sin2x+cos2x=√2(sin2x*√2/2+cos2x*√2/2)这步是提取根号2= √2(sin2xcosπ/4+cos2xsinπ/4)=√2sin(2x+π/4)