Question

化简(1-cos^4α-sin^4α)/(1-cos^6α-sin^6α)

Answer 1

你好！

1-cos^6α-sin^6α
=1- [(cos²a)³+(sin²a)³]
=1- (cos²a+sin²a)(cos⁴a - cos²asin²a +sin⁴a)
=1 - cos⁴a -sin⁴a +cos²asin²a

1 - cos⁴a -sin⁴a
=sin²a +cos²a - cos⁴a - sin⁴a
=sin²a(1-sin²a) + cos²a(1-cos²a)
=sin²acos²a +cos²asin²a
=2sin²acos²a

∴原式= 2sin²acos²a / (2sin²acos²a +cos²asin²a) = 2/3

Answer 2

(1-cos^4α-sin^4α)/(1-cos^6α-sin^6α)
=(sin^2α+cos^2α-cos^4α-sin^4α)/[1-(cos^6α+sin^6α)]
=[sin^2α(1-sin^2α)+cos^2α(1-cos^2α]/[1-(cos^2α+sin^2α)(cos^4α-cos^2αsin^2α+sin^4α)]
=[sin^2αcos^2α+sin^2αcos^2α]/[1-(cos^4α-cos^2αsin^2α+sin^4α)]
=2sin^2αcos^2α/[sin^2α+cos^2α-cos^4α+cos^2αsin^2α-sin^4α]
=2sin^2αcos^2α/[sin^2α-sin^4α+cos^2α-cos^4α+cos^2αsin^2α]
=2sin^2αcos^2α/[sin^2α(1-sin^2α)+cos^2α(1-cos^2α)+cos^2αsin^2α]
=2sin^2αcos^2α/[sin^2αcos^2α+cos^2αsin^2α+cos^2αsin^2α]
=2sin^2αcos^2α/[3sin^2αcos^2α]
=2/3

Answer 3

1-(cosα)^4-(sinα)^4=(sinα)^2+(cosα)^2-(cosα)^4-(sinα)^4
=(sinα)^2*[1-(sinα)^2] +(cosα)^2[1-(cosα)^2]
=2(sinα)^2*(cosα)^2

1-(cosα)^6-(sinα)^6=(sinα)^2+(cosα)^2-(cosα)^6-(sinα)^6
=(sinα)^2*[1-(sinα)^4] +(cosα)^2[1-(cosα)^4]
=(sinα)^2*[(sinα)^2+(cosα)^2-(sinα)^4] +(cosα)^2[(sinα)^2+(cosα)^2-(cosα)^4]
=(sinα)^2*[(sinα)^2(1-(sinα)^2)+(cosα)^2] +(cosα)^2[(sinα)^2+(cosα)^2(1-(cosα)^2)]
=(sinα)^2*[(sinα)^2(cosα)^2+(cosα)^2] +(cosα)^2[(sinα)^2+(cosα)^2(sinα)^2]
=(sinα)^2*(cosα)^2*[(sinα)^2+1] +(cosα)^2*(sinα)^2*[1+(cosα)^2]
=(sinα)^2*(cosα)^2*[(sinα)^2+1+1+(cosα)^2]
=3*sinα)^2*(cosα)^2

(1-cos^4α-sin^4α)/(1-cos^6α-sin^6α)=2/3

Answer 4

ans =

2/3

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