作业帮(sin^6)x+(sin^6)x怎么求周期呀?
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(sin^6)x+(sin^6)x怎么求周期呀?
(sin^6)x+(sin^6)x怎么求周期呀?
(sin^6)x+(sin^6)x怎么求周期呀?
题目有歧义
具体点看不懂
还是派
题有错，是不是这样y=(sin^6)x+(cos^6)x
y=(sin^6)x+(cos^6)x
=(sin^2x+cos^2x)^3-3sin^2xcos^2x(sin^2x+cos^2x)
=1-3sin^2xcos^2x
=1-3/4(1-cos2x)(1+cos2x)
=1-3/4[1-cos^2(2x)]
=1/4+3/4cos^2(2x)...
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题有错，是不是这样y=(sin^6)x+(cos^6)x
y=(sin^6)x+(cos^6)x
=(sin^2x+cos^2x)^3-3sin^2xcos^2x(sin^2x+cos^2x)
=1-3sin^2xcos^2x
=1-3/4(1-cos2x)(1+cos2x)
=1-3/4[1-cos^2(2x)]
=1/4+3/4cos^2(2x)
=1/4+3/8(1+cos4x)
=3/8cos4x+5/8
T=2π/4=π/2
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