sin^4-cos^4的最小正周期是我错在哪?f(x)=(sinx)^4+(cosx)^4=[(sinx)^2+(cosx)^2]^2 -2(sinx)^2(cosx)^2=1 -[sin(2x)]^2/2=1-[1-cos(4x)]/4=cos(4x)/4 +3/42π/4=π/2函数的最小正周期为π/2

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sin^4-cos^4的最小正周期是我错在哪?f(x)=(sinx)^4+(cosx)^4=[(sinx)^2+(cosx)^2]^2 -2(sinx)^2(cosx)^2=1 -[sin(2x)]^2/2=1-[1-cos(4x)]/4=cos(4x)/4 +3/42π/4=π/2函数的最小正周期为π/2
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sin^4-cos^4的最小正周期是我错在哪?f(x)=(sinx)^4+(cosx)^4=[(sinx)^2+(cosx)^2]^2 -2(sinx)^2(cosx)^2=1 -[sin(2x)]^2/2=1-[1-cos(4x)]/4=cos(4x)/4 +3/42π/4=π/2函数的最小正周期为π/2
sin^4-cos^4的最小正周期是
我错在哪?f(x)=(sinx)^4+(cosx)^4
=[(sinx)^2+(cosx)^2]^2 -2(sinx)^2(cosx)^2
=1 -[sin(2x)]^2/2
=1-[1-cos(4x)]/4
=cos(4x)/4 +3/4
2π/4=π/2
函数的最小正周期为π/2

sin^4-cos^4的最小正周期是我错在哪?f(x)=(sinx)^4+(cosx)^4=[(sinx)^2+(cosx)^2]^2 -2(sinx)^2(cosx)^2=1 -[sin(2x)]^2/2=1-[1-cos(4x)]/4=cos(4x)/4 +3/42π/4=π/2函数的最小正周期为π/2
是 f(x)=(sinx)^4+(cosx)^4
还是f(x)=(sinx)^4-(cosx)^4
如果是f(x)=(sinx)^4+(cosx)^4,你的答案没错
如果是f(x)=(sinx)^4-(cosx)^4
=[(sinx)^2+(cosx)^2][(sinx)^2-(cosx)^2]
=-cos2x
2π/2=π