求解一道数学题

Question

求值，题目在图片中。

Answer 1

∵1=sin^2α+cos^2α
∴
原式=(sin^2α+cos^2α-cos^6α-sin^6α) /(sin^2α+cos^2α-cos^4α-sin^4α)
分母=sin^2α(1-sin^2α)+cos^2α(1-cos^2α)=2sin^2αcos^2α
分子=sin^2α(1-sin^4α)+cos^2α(1-cos^4α)
∵sin^2α(1-sin^4α)=sin^2α(1-sin^2α)(1+sin^2α)=sin^2αcos^2α(1+sin^2α)
∴同理cos^2α(1-cos^4α)=sin^2αcos^2α(1+cos^2α)
∴
原式
=sin^2αcos^2α(1+sin^2α+1+cos^2α)/2sin^2αcos^2α
=3/2

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Answer 2

∵1=sin^2α+cos^2α
∴
原式=(sin^2α+cos^2α-cos^6α-sin^6α) /(sin^2α+cos^2α-cos^4α-sin^4α)
分母=sin^2α(1-sin^2α)+cos^2α(1-cos^2α)=2sin^2αcos^2α
分子=sin^2α(1-sin^4α)+cos^2α(1-cos^4α)
∵sin^2α(1-sin^4α)=sin^2α(1-sin^2α)(1+sin^2α)=sin^2αcos^2α(1+sin^2α)
∴同理cos^2α(1-cos^4α)=sin^2αcos^2α(1+cos^2α)
∴
原式
=sin^2αcos^2α(1+sin^2α+1+cos^2α)/2sin^2αcos^2α
=3/2

Answer 3

降幂嘛，用二倍角公式，答案是1.5.

Answer 4

∵1=sin^2α+cos^2α
∴
原式=(sin^2α+cos^2α-cos^6α-sin^6α) /(sin^2α+cos^2α-cos^4α-sin^4α)
分母=sin^2α(1-sin^2α)+cos^2α(1-cos^2α)=2sin^2αcos^2α
分子=sin^2α(1-sin^4α)+cos^2α(1-cos^4α)
∵sin^2α(1-sin^4α)=sin^2α(1-sin^2α)(1+sin^2α)=sin^2αcos^2α(1+sin^2α)
∴同理cos^2α(1-cos^4α)=sin^2αcos^2α(1+cos^2α)
∴
原式
=sin^2αcos^2α(1+sin^2α+1+cos^2α)/2sin^2αcos^2α
=3/2
大概应该是这样

Answer 5

这些是神马？

Answer 6

3/2

参考资料：http://zhidao.baidu.com/question/304457686.html?push=being