Question

已知函数f(x)=sin^2wx+√3sinwxsin(wx+π/2)(w>0)的最小正周期为π

Answer 1

①f(x)=sin^2wx+√3sinwxsin(wx+π/2)
=1/2(1-cos2wx)
+√3sinwx
coswx
=1/2(1-cos2wx)
+√3/2sin2wx
=√3/2sin2wx-1/2cos2wx+1/2
=sin(2wx-π/6)
+1/2
最小正周期为π
所以w=1，f(x)=
sin(2x-π/6)
+1/2。
所以函数值域是[-1/2,3/2].
②x∈[-π/12,π/2]时，f(x)=
sin(2x-π/6)
+1/2
则2x-π/6∈[-π/3,5π/6]
sin(2x-π/6)∈[-√3/2,1].
所以函数值域是[(-√3+1)/2,3/2].

Answer 2

f(x)=(1-cos2wx)/2+√3/2*sin2x
=(√3/2)sin2wx-1/2*cos2wx+1/2
=√[(√3/2)^2+(1/2)^2]*sin(2wx-z)+1/2
其中tanz=(1/2)/(√3/2)
所以z=π/6
所以f(x)=sin(2wx-π/6/6)+1/2
T=2π/|2w|=π
w>0
所以w=1
f(x)=sin(2x-π/6)+1/2
0<=x<=2π/3
-π/6<=2x-π/6<=7π/6
sin在(2kπ-π/2,2kπ+π/2)是
增函数

在(2kπ+π/2,2kπ-π/2)是
减函数

所以-π/6<=2x-π/6<=π/2
-1/2<=sin(2x-π/6)<=1
π/2<=2x-π/6<=7π/6
-1/2<=sin(2x-π/6)<=1
所以-1/2<=sin(2x-π/6)<=1
0<=sin(2x-π/6)+1/2<=3/2
所以f(x)在区间[0,2π/3]上的取值是[0,3/2]