分类：教育/科学-学习帮助
问题：高一数学题求助,大家快来,明早要交~已知向量a=(2cos0.5 
已知向量a=(2c0.5x,tan(x/2+π/4)),向量b=(√2*sin(x/2+π/4),tan(x/2-π/4)),令f(x)=a*b.求函数f(x)的最大值,最小正周期,并写出f(x)在[0,π]上的单调区间.
回答：解：f(x)=a*b=2cos0.5x×√2*sin(x/2+π/4)+tan(x/2+π/4)×tan(x/2-π/4)
=2cos0.5x×√2×[sin0.5xcos(π/4)+cos0.5xsin(π/4)+tan(π/2+x/2-π/4)×tan(x/2-π/4)
=2cos0.5x×(sin0.5x+cos0.5x)-cot(x/2-π/4)×tan(x/2-π/4)
=sinx+2(cos0.5x)＾-1
=sinx+cosx=√2[sinxcos(π/4)+cosxsin(π/4)]
=√2sin(x+π/4)
[f(x)]max=√2
最小正周期   2π
x∈[0,π/4]   f(x)单调递增 
x∈(π/4,π]   f(x)单调递减.