(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]
化簡:
化簡:
(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)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)
數(shù)學(xué)人氣:995 ℃時間:2020-06-25 03:02:28
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