Question

函数y=sin^4-cos^4的最小正周期是多少？ 要答案和步骤，谢谢！

Answer 1

y=sin^4x-cos^4x
=(sin²x+cos²x)(sin²x-cos²x)
=sin²x-cos²x
=-2cos2x
最小正周期T=2π/2=π
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Answer 2

y =sin^4α-cos^4α
=（sin^2α+cos^2α）（sin^2α-cos^2α）
= -（cos^2α-sin^2α）
= -cos（2α）
所以：函数y=sin^4-cos^4的最小正周期是π

Answer 3

是y=(sinx)^4-(cosx)^4吗
y=(sinx)^4-(cosx)^4
=[(sinx)^2+(cosx)^2][(sinx)^2-(cosx)^2]
=(sinx)^2-(cosx)^2
=(1-cos2x)/2-(1+cos2x)/2
=-cos2x
cos2x的周期是π
所以y=(sinx)^4-(cosx)^4最小正周期π

Answer 4

y==sin^4-cos^4=（sin^2-cos^2）（sin^2+cos^2）=sin^2-cos^2=-cos2x
故最小正周期为π

Answer 5

y=sinx^4-cosx^4
=(sinx^2-cosx^2)(sinx^2+cosx^2)
=sinx^2-cosx^2
=sinx^2-(1-sinx^2)
=-(1-2sinx^2)
=-cos2x
知，最小正周期
T=2π /2=π