高一数学，要过程。1.求f(x)=CosX+Cos(x+兀/3)的最值2.X属于[-兀/2，兀/2]

显示全部楼层 · 2013-1-7 11:37:46

f(x)=CosX+Cos(x+兀/3)
=cosx+cosxcosπ/3-sinxsinπ/3
=3/2cosx-√3/2sinx =√3(√3/2cosx-1/2sinx) =√3cos(x+π/6) 最大值为√3，最小值为-√3（2）f(X)=SinX+根号Cosx
=2（1/2sinx+√3/2cosx)
=2sin(x+π/3)∵x∈[-π/2,π/2]∴x+π/3∈[-π/6,5π/6]∴sin(x+π/3)∈[-1/2,1]∴2sin(x+π/3)∈[-1，2]即函数最大值为2，最小值-1...

千问 · 2013-1-7 11:37:46

解：1、∵f(x)=cosx+cos(x+π/3)
=cosx+[(cosx)/2]-[(√3sinx)/2]
=√3[(√3/2)cosx-(1/2)sinx]
=√3cos(x+π/6)∴当x=-π/6时，f(x)=√3...

千问 · 2013-1-7 11:37:46

f(x)=cosx+(cosxcosπ/3-sinxsinπ/3)=cosx+(cosx*1/2-sinx*√3/2)=cosx*3/2-sinx*√3/2=√3(cosx*√3/2-sinx*1/2)=√3(cosxcosπ/6-sinxsinπ/6)=√3cos(x+π/6)max f（x）=√3，min f(x)=√-3...