求助,这道题怎么做,第一步怎么化简?
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(1)
f(x)=cos(2x- π/3)+2sin(x- π/4)sin(x+ π/4)
=cos2xcos(π/3)+sin2xsin(π/3)+cos2x-cos(-π/2)
=½cos2x+(√3/2)sin2x+cos2x-0
=(√3/2)sin2x+(3/2)cos2x
=√3[(1/2)sin2x+(√3/2)cos2x]
=√3sin(2x+π/3)
最小正周期T=2π/2=π
f(x)=√3sin(2x+π/3)=√3sin[2(x+π/6)]
π/2- π/6=π/3
对称轴方程:x=kπ+ π/3,(k∈Z)
(2)
x∈[-π/12,π/2]
-π/2≤2x+π/3≤2π/3
-1≤sin(2x+π/3)≤1
-√3≤√3sin(2x+π/3)≤√3
f(x)的值域为[-√3,√3]
f(x)=cos(2x- π/3)+2sin(x- π/4)sin(x+ π/4)
=cos2xcos(π/3)+sin2xsin(π/3)+cos2x-cos(-π/2)
=½cos2x+(√3/2)sin2x+cos2x-0
=(√3/2)sin2x+(3/2)cos2x
=√3[(1/2)sin2x+(√3/2)cos2x]
=√3sin(2x+π/3)
最小正周期T=2π/2=π
f(x)=√3sin(2x+π/3)=√3sin[2(x+π/6)]
π/2- π/6=π/3
对称轴方程:x=kπ+ π/3,(k∈Z)
(2)
x∈[-π/12,π/2]
-π/2≤2x+π/3≤2π/3
-1≤sin(2x+π/3)≤1
-√3≤√3sin(2x+π/3)≤√3
f(x)的值域为[-√3,√3]
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