作业帮已知函数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](/uploads/image/z/15255139-67-9.jpg?t=%E5%B7%B2%E7%9F%A5%E5%87%BD%E6%95%B0f%28x%29%3Dsin%5E2%28x-%CF%80%2F6%29%2Bsin%5E2%28x%2B%CF%80%2F6%29%2C%E8%8B%A5x%E2%88%88%5B-%CF%80%2F3%2C%CF%80%2F6%5D%2C%E6%B1%82%E5%87%BD%E6%95%B0f%28x%29%E7%9A%84%E5%80%BC%E5%9F%9F)
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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