(1-sin^6θ-cos^6θ)/(1-sin^4θ-cos^4θ)

长空雁过 1年前 已收到1个回答 举报

relaxheng 幼苗

共回答了21个问题采纳率:71.4% 举报

1-sin^6θ-cos^6θ
=1-[(sin^2θ)^3+(cos^2θ)^3]
=1-(sin^2θ+cos^2θ)(sin^4θ-sin^2θcos^2θ+cos^4θ)
=1-(sin^4θ-sin^2θcos^2θ+cos^4θ)
=1-(sin^4θ+2sin^2θcos^2θ+cos^4θ-3sin^2θcos^2θ)
=1-[(sin^2θ+cos^2θ)^2-3sin^2θcos^2θ]
=3sin^2θcos^2θ
1-sin^4θ-cos^4θ
=1-(sin^4θ+cos^4θ)
=1-[(sin^4θ+2sin^2θcos^2θ+cos^4θ-2sin^2θcos^2θ)]
=1-[(sin^2θ+cos^2θ)^2-2sin^2θcos^2θ]
=2sin^2θcos^2θ
所以:(1-sin^6θ-cos^6θ)/(1-sin^4θ-cos^4θ)
=(3sin^2θcos^2θ)/(2sin^2θcos^2θ)
=3/2.

1年前

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