解: y=cosx(sinx+cosx)
=sinxcosx + (cosx)^2
=sin2x /2 + (cos2x + 1)/2
=(sin2x + cos2x)/2 + 1/伏明碧2
=(√2 /2 )sin(2x + π/4) + 1/槐槐2
所以最小正周期=2π /缺举2=π
PI
y=cosxsinx+(卖纤派cosx)^2
=1/2sin2x+1/2(cos2x+1)
=1/2+1/2(sin2x+cos2x)
=1/2+√2/2sin(2x+π/4)
因此,周期中贺为π
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