已知向量a=(cosx/2,tan(x/2+π/4)),向量b=(√2sin(x/2+π/4),tan(x/2-π/4)),令f(x)=ab
问题描述:
已知向量a=(cosx/2,tan(x/2+π/4)),向量b=(√2sin(x/2+π/4),tan(x/2-π/4)),令f(x)=ab
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
答
f(x)=2cosx/2×(√2sin(x/2+π/4)+ tan(x/2+π/4)×tan(x/2-π/4)) =√2[sin(x+π/4)+sin(π/4)] + [1+tan(x/2)]/[1-tan(x/2)]×[tan(x/2)-1]/[1+tan(x/2)] =√2sin(x+π/4) 最大值=√2最小正周期=2πsinx的增区...