求函数f(x)=sin^6+cos^6的周期和最值

f(x)=sin^6x+cos^6x
=(sin²x)³+(cos²x)³
=(sin²x+cos²x)(sin⁴x-sin²xcos²x+cos⁴x)
=1*(sin⁴x+2sin²xcos²x+cos⁴x-3sin²xcos²x)
=(sin²x+cos²x)²-3sin²xcos²x
=1²-3sin²xcos²x
=-3(sinxcosx)²x+1
=-3(½sin2x)²+1
=-3/4sin²2x+1
=-3/4*(1-cos4x)/2+1
=3/8cos4x+5/8
周期T=2π/ω=2π/4=π/2
当cos4x=1时取得最大值3/8+5/8=1
当cos4x=-1时取得最小值-3/8+5/8=1/4

f(x)=sin^6x+cos^6x=(sin^2x+cos^x)(sin^4x-sin^2x*cos^2x+cos^4x)
=sin^4x-sin^2x*cos^2x+cos^4x=sin^4x+2sin^x*cos^2x+cos^4x-3sin^2x*cos^2x
=(sin^2x+cos^2x)^2-3/4sin^2(2x)
=1-3/8(1-cos4x)
=3/8*cos4x+5/8
周期T=2π/w=2π/4=π/2
最大值 1 当cos4x=1时
最小值 1/4当cos4x=-1时

f(x)=(sin²x+cos²x)(sin⁴x-sin²xcos²x+cos⁴4)=1*[(sin²x+cos²x)²-3sin²xcos²x]=1²-3/4*(2sinxcosx)²=1-3/4*sin²2x=1-3/4*(1-cos4x)/2=5/8...

你好，很高兴回答你的问题
①周期
sin^6(x)+cos^6(x)=[sin^2(x)]^3+[cos^2(x)}^3
=[sin^2(x)+cos^2(x)}*[sin^4(x)]+[cos^4(x)-sin^2(x)cos^2}
=1*[sin^4(x)]^3+[cos^4(x)-sin^2(x)cos^2}
=[sin^2(x)]^+cos^^2(x)]^2-3sin^2(x)cos^2}
=1-3sin^2(x)cos^2}
=1-3/4(sin2x)^2
=1-(3/4)*(1-cos4x)/2
最小正周期是T=2π/4=π/2

②最值
f(x)=sin^6+cos^6
f(x)=(sin^2 x+cos^2 x)^3-3sin^4 xcos^2 x-3sin^2 xcos^4 x
=1-3sin^2 xcos^2 x(sin^2 x+cos^2 x)
=1-3/4*sin^2 4x
=1-3/4*(1-cos8x)/2
=5/8+3/8*cos8x
最大值：f（x）=1
最小值：f（x）=1/4