Question

已知函数fx=2sinxsin(x+π/6) 求最小正周期单调增区间 x属于【0，π/2】 求值域

Answer 1

解fx=2sinxsin(x+π/6)
=cos[x-(x+π/6)]-cos[x+(x+π/6)]
=-cos(2x+π/6)+√3/2
故函数的周期T=2π/2=π
当2kπ≤2x+π/6≤2kπ+π，k属于Z函数是增函数
故函数的增区间是[kπ-π/12,kπ+5π/12],k属于Z
由x属于【0，π/2】
知2x属于【0，π】
即2x+π/6属于【π/6，7π/6】
即cos(2x+π/6)属于[-1,1/2]
即-cos(2x+π/6)属于[-1/2,1]
即-cos(2x+π/6)+√3/2属于[-1/2+√3/2,1+√3/2]
故函数的值域为[-1/2+√3/2,1+√3/2]

Answer 2

f(x)=2sinxsin(x+π/6)
=2sinx(√3/2+1/2cosx)
=√3(sinx)^2+sinxcosx
=√3/2(1-cos2x)+1/2sin2x
=1/2sin2x-√3/2cos2x+√3/2
=sin(2x-π/3)+√3/2
T=2π/2=π
x属于【0，π/2】
2x-π/3属于【-π/3,2π/3】
sin(2x-π/3)在【-π/3,2π/3】的值域为:[-√3/2,1]
f(x)=sin(2x-π/3)+√3/2的值域为:[0,(2+√3)/2]