Question

已知函数f(X)=sin的平方(x--π/6）+cos的平方（x--π/3)+sinx.cosx,x属于R　　　求f(x)的最大值及取得最大

Answer 1

已知函数f(X)=sin²(x-π/6)+cos²(x-π/3)+sinxcosx，x属于R；求f(x)的最大值及取得最大值时的x值
解：f(X)=sin²(x-π/6)+cos²(x-π/3)+sinxcosx
=[(√3/2)sinx-(1/2)cosx]²+[(1/2)cosx+(√3/2)sinx]+(1/2)sin2x
=[(3/4)sin²x-(√3/4)sinxcosx+(1/4)cos²x]+[(1/4)cos²x+(√3/4)sinxcosx+(3/4)sin²x]+(1/2)sin2x
=(3/2)sin²x+(1/2)cos²x+(1/2)sin2x
=(3/4)(1-cos2x)+(1/4)(1+cos2x)+(1/2)sin2x
=1+(1/2)(sin2x-cos2x)=1+(√2/2)sin(2x-π/4)
故(2-√2)/2≦f(x)≦(2+√2)/2，当2x-π/4=π/2+2kπ，即x=3π/8+kπ (k∈R)是f(x)获得最大值
(2+√2)/2.

Answer 2

请看

Answer 3

2f(x)=2sin²(x-π/6)+2cos²(x-π/3)+2sinxcosx
=1-cos(2x-π/3)+cos(2x-2π/3)+1+sin2x
=2sin[(2x-2π/3+2x-π/3)/2]sin[(2x-2π/3-2x+π/3)/2]+sin2x
=2cos2xsinπ/6+sin2x
=cos2x+sin2x
=√2sin(2x+π/4)
当sin(2x+π/4)=1，即x=kπ+π/8时，f(x)最大=(√2)/2