三角函数求救

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三角函数求救
 

f(x)=sin(x+π/3)sin(x+π/2)
=sin(x+π/3)cosx
=1/2sinxcosx+√3/2cos²x
=1/4sin2x+√3/4(2cos²x-1)+√3/4
=1/2(1/2sin2x+√3/2cos2x)+√3/4
=1/2(cosπ/3sin2x+sinπ/3cos2x)+√3/4
=1/2sin(2x+π/3)+√3/4
∴T=2π/2=π
(2)g(x)=f(x)-√3/4=1/2sin(2x+π/3)
x∈[0,π/2]
2x+π/3∈[π/3,π+π/3]
∴g(x)∈[-√3/2,1]