作业帮求函数y=sinx^4+2*根号3*sinxcosx-cosx^4最小正周期和最小值,并写出该函数在[0,π]上的单调区间
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![求函数y=sinx^4+2*根号3*sinxcosx-cosx^4最小正周期和最小值,并写出该函数在[0,π]上的单调区间](/uploads/image/z/1043680-40-0.jpg?t=%E6%B1%82%E5%87%BD%E6%95%B0y%3Dsinx%5E4%2B2%2A%E6%A0%B9%E5%8F%B73%2Asinxcosx-cosx%5E4%E6%9C%80%E5%B0%8F%E6%AD%A3%E5%91%A8%E6%9C%9F%E5%92%8C%E6%9C%80%E5%B0%8F%E5%80%BC%2C%E5%B9%B6%E5%86%99%E5%87%BA%E8%AF%A5%E5%87%BD%E6%95%B0%E5%9C%A8%5B0%2C%CF%80%5D%E4%B8%8A%E7%9A%84%E5%8D%95%E8%B0%83%E5%8C%BA%E9%97%B4)
求函数y=sinx^4+2*根号3*sinxcosx-cosx^4最小正周期和最小值,并写出该函数在[0,π]上的单调区间
求函数y=sinx^4+2*根号3*sinxcosx-cosx^4最小正周期和最小值,并写出该函数在[0,π]上的单调区间
求函数y=sinx^4+2*根号3*sinxcosx-cosx^4最小正周期和最小值,并写出该函数在[0,π]上的单调区间
y=sinx^4-cosx^4+2*√3*sinxcosx
=(sinx^2+cosx^2)(sinx^2-cosx^2)+√3sin2x
=√3sin2x+(sinx^2-cosx^2)
=√3sin2x-cos2x
=2sin(2x-π/6)
最小正周期 2π/2=π
最小值 -2
y=2sin(2x-π/6)的单调增区间[kπ-π/6,kπ+π/3]
单调减区间[kπ+π/3,kπ+5π/6]
故该函数在[0,π]上的单调增区间[0,π/3]和[5π/6,π]
单调减区间[π/3,5π/6]
y=(sinx^4 -cosx^4)+2√3*sinxcosx
=(sin^2 x -cos^2 x)(sin^2 x +cos^2 x)+√3sin2x
=(sin^2 x -cos^2 x)*1+√3sin2x
=cos2x +√3sin2x
=2[(1/2)*cos2x +(√3/2)*sin2x]
=2*[sin(π/6)*cos2x +cos(π...
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y=(sinx^4 -cosx^4)+2√3*sinxcosx
=(sin^2 x -cos^2 x)(sin^2 x +cos^2 x)+√3sin2x
=(sin^2 x -cos^2 x)*1+√3sin2x
=cos2x +√3sin2x
=2[(1/2)*cos2x +(√3/2)*sin2x]
=2*[sin(π/6)*cos2x +cos(π/6)*sin2x]
=2*sin(2x + π/6)
则最小正周期是 2π/2=π;
最小值是-2;
x∈[0，π],则2x+π/6∈[π/6,13π/6];
则当2x+π/6=π/2即当x=π/6时,y取得最大值2;
由三角函数的性质知,当x∈[0，π/6]时,函数单调增加;
当x∈[π/6,π]时,函数单调递减
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