Question

已知函数f(x)=sinx(sinx+根号三cosX),其中x属于【0,π/2】

1.求 f(x)的最大值和最小值
2.若 cos(阿尔法+π/6）=3/4，求f(阿尔法）的值
求详解啊。谢谢~~~

Answer 1

f(x)=sin²x＋√3sinxcosx
=(1/2)(1－cos2x)＋(√3/2)sin2x
=√3/2)sin2x－(1/2)cos2x＋1/2
=sin(2x－π/6)＋1/2
x∈【0,π/2】所以 2x-π/6∈[-π/6,5π/6]
根据函数图象可知
所以f(x)最大值为3/2 此时x=π/3
f(x)最小值为0 此时x=0
f(α)=sin（2α-π/6）+1/2
=cos(2α-π/6+π/2)+1/2
=cos(2α+π/3)+1/2
=2cos²(α+π/6）-1+1/2
=2*9/16-1/2
=5/8

Answer 2

1、
f(x)=sinx(sinx+√3cosx)
=sin²x + √3sinxcosx
= (1 - cos2x)/2 + (√3/2)sin2x
=(√3/2)sin2x - (1/2)cos2x + 1/2
=sin(2x - π/6) + 1/2
0≤x≤π/2
0≤2x≤π
-π/6≤2x - π/6≤5π/6
-1/2≤sin(2x - π/6)≤1
0≤sin(2x - π/6) + 1/2≤3/2
∴f(x)最大值、最小值分别为3/2、 0

2、
cos(2α + π/3) = cos[2(α+π/6)] = 2cos²(α+π/6) - 1=1/8
∴f(α)= sin(2α - π/6) + 1/2
=sin [(2α + π/3) - π/2] + 1/2
= - cos(2α + π/3) + 1/2
= 3/8

Answer 3

f(x)=sin²x＋√3sinxcosx
=(1/2)(1－cos2x)＋(√3/2)sin2x
=(√3/2)sin2x－(1/2)cos2x＋(1/2)
=sin(2x－π/6)＋(1/2)
则最大值是3/2，最小值是－(1/2)

cos(a＋π/6)=3/4，f(a)=sin(2a－π/6)＋(1/2)=－cos(2a＋π/3)＋(1/2)
=－cos[2(a＋π/6)]＋(1/2)=－[2cos²(a＋π/6)－1]＋(1/2)
=－2cos²(a＋π/6)＋(3/2)=3/8