已知f[x]=[sinx+cosx]的平方+cos2x
【1】求f【x】的最小正周期
【2】f[x]的区间【0,π/2】上最大值和最小值
f(x)=(sinx+cosx)²+cos2x
f(x)=1+sin2x+cos2x
f(x)=1+√2·sin(2x+π/4)
f(x)的最小正周期=2π/2=π
2x+π/4=π/2,即:x=π/8时,sin(2x+π/4)=1,f(x)有最大值1+√2
2x+π/4=3π/2,即:x=5π/8时,sin(2x+π/4)=-1,f(x)有最小值1-√2
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