化简:(1)sin^2·α+cos^4·α+sin^2·αc-查字典问答网
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  化简:(1)sin^2·α+cos^4·α+sin^2·αcos^2·α(2)(1-cos^4·α-sin^4·α)/(1-cos^6·α-sin^4·α)

  化简:

  (1)sin^2·α+cos^4·α+sin^2·αcos^2·α

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

1回答
2020-09-2421:13
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马树萍

  (1)sin^2·α+cos^4·α+sin^2·αcos^2·α

  =sin^2·α+cos^2·α(cos^2·α+sin^2·α)

  =sin^2·α+cos^2·α

  =1

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

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

  =[cos^2·α+cos^2·αsin^2·α-cos^4·α]/[cos^2·α+cos^2·αsin^2·α-cos^6·α]

  =[cos^2·α(1-cos^2·α)+cos^2·αsin^2·α]/[cos^2·α(1-cos^4·α)+cos^2·αsin^2·α]

  =[2cos^2·αsin^2·α]/[cos^2·α(1-cos^2·α)(1+cos^2·α)+cos^2·αsin^2·α]

  =[2cos^2·αsin^2·α]/[cos^2·αsin^2·α(1+cos^2·α)+cos^2·αsin^2·α]

  =[2cos^2·αsin^2·α]/[2cos^2·αsin^2·α+sin^2·αcos^4·α]

  =2/[2+cos^2·α]

2020-09-24 21:18:12
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