(1-sin^6 a-cos^6 a)/(1-sin^4 a-cos^4 a)的化简结果
来源:学生作业帮 编辑:作业帮 分类:数学作业 时间:2024/07/14 03:33:28
(1-sin^6 a-cos^6 a)/(1-sin^4 a-cos^4 a)的化简结果
(1-sin^6 a-cos^6 a)/(1-sin^4 a-cos^4 a)
=[1-(sin^6 a+cos^6 a)]/[(1-sin^4 a-cos^4 a]
=[1-(sin^2 a+cos^2 a)(sin^4 a-sin^2 acos^2 a+cos^4 a)]/[(1-sin^4 a-cos^4 a]
=[1-(sin^4 a-sin^2 acos^2 a+cos^4 a)]/(1-sin^4 a-cos^4 a)
=[1-sin^4 a+sin^2 acos^2 a-cos^4 a]/(1-sin^4 a-cos^4 a)
=1+sin^2 acos^2 a/(1-sin^4 a-cos^4 a)
=1+sin^2 acos^2 a/[(1-sin^4 a)-cos^4 a]
=1+sin^2 acos^2 a/[(1-sin^2 a)(1+sin^2 a)-cos^4 a]
=1+sin^2 acos^2 a/[cos^2 a(1+sin^2 a)-cos^4 a]
=1+sin^2 acos^2 a/[cos^2 a(1+sin^2 a-cos^2a)]
=1+sin^2 acos^2 a/[2sin^2 a*cos^2 a]
=1+1/2
=3/2
=[1-(sin^6 a+cos^6 a)]/[(1-sin^4 a-cos^4 a]
=[1-(sin^2 a+cos^2 a)(sin^4 a-sin^2 acos^2 a+cos^4 a)]/[(1-sin^4 a-cos^4 a]
=[1-(sin^4 a-sin^2 acos^2 a+cos^4 a)]/(1-sin^4 a-cos^4 a)
=[1-sin^4 a+sin^2 acos^2 a-cos^4 a]/(1-sin^4 a-cos^4 a)
=1+sin^2 acos^2 a/(1-sin^4 a-cos^4 a)
=1+sin^2 acos^2 a/[(1-sin^4 a)-cos^4 a]
=1+sin^2 acos^2 a/[(1-sin^2 a)(1+sin^2 a)-cos^4 a]
=1+sin^2 acos^2 a/[cos^2 a(1+sin^2 a)-cos^4 a]
=1+sin^2 acos^2 a/[cos^2 a(1+sin^2 a-cos^2a)]
=1+sin^2 acos^2 a/[2sin^2 a*cos^2 a]
=1+1/2
=3/2
(1-sin^6 a-cos^6 a)/(1-sin^4 a-cos^4 a)的化简结果
1-sin^6a-cos^6a分之1-sin^4a-cos^4a 化简
求值:1-sin^6a-cos^6a/1-sin^4a-cos^4a
化简1-sin^4a-cos^4a/1-sin^6a-cos^6a
(1-sin^4a-cos^4a)/(1-sin^6a-cos^6a)
(1-sin^6a-cos^6a)/(1-sin^4a-cos^4a)=?
化简(1-sin^6 a-cos^6 a)/(cos^2 a-cos^4 a)==
(1-sin^6a-cos^6a)/(sin^2a-sin^4a)
1-cos^6(a)-sin^6(a)化简得3sin^(a)cos^(a)的详细过程!
化简 1-sin^4a-cos^4a/cos^2a-cos^4a
证明sin(4A)sin(2A)(1-cos(2A)) cos(4A)cos(2A)(1 cos(2A))=cos(2A
若sin^4a/sin^2b+cos^4a/cos^2b=1,证明sin^4b/sin^2a+cos^4b/cos^2a