已知函数f(x)=sin^2(x-π/6)+sin^2(x+π/6),若x∈[-π/3,π/6],求函数f(x)的值域
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已知函数f(x)=sin^2(x-π/6)+sin^2(x+π/6),若x∈[-π/3,π/6],求函数f(x)的值域
已知函数f(x)=sin^2(x-π/6)+sin^2(x+π/6),若x∈[-π/3,π/6],求函数f(x)的值域
已知函数f(x)=sin^2(x-π/6)+sin^2(x+π/6),若x∈[-π/3,π/6],求函数f(x)的值域
f(x)=sin^2(x-π/6)+sin^2(x+π/6)
=(1/2)[1-cos2(x-π/6)+1-cos2(x+π/6)]
=(1/2)[2-2cos2xcos(π/3)]
=1-(1/2)cos2x
2x∈[-2π/3,π/3]
所以cos2x∈[-1/2,1]
所以f(x)∈[1/2,5/4]
f(x)=sin²(x-π/6)+sin²(x+π/6)
=[1-cos(2x-π/3)]/2+[1-cos(2x+π/3)]/2
=1-[cos(2x+π/3)+cos(2x-π/3)]/2
=1-cos2xcos(π/3)
=1-(1/2)cos2x.
∵-1≤cos2x≤1,
∴1/2≤f(x)≤3/2,
故函数值域为:[1/2, 3/2]。
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