三角函数——数学

来源：百度知道 编辑：UC知道 时间：2024/06/07 06:05:13
1.求1-sin^6x-cos^6x/1-sin^4x-sin^4x
注意：^n代表的是几次幂

请写详细过程，谢谢o(∩_∩)o...

题目应该是
1-sin^6 x-cos^6 x
------------------
1-sin^4x-cos^4x
=1-(sin^2 x +cos^2 x)(sin^4 x +sin^2 x cos^2x +cos^4 x)
-------------------------------------------------
1-(sin^2 x+cos^2 x)(sin^2 x-cos^2 x)

=1-[(sin^2 x +cos^2 x)^2-3sin^2 x cos^2x ]
-----------------------------------------
1-sin^2 x+cos^2 x

=3sin^2 x cos^2x /2cos^2 x
=
3/2sin^2 x

（1-sin^6x-cos^6x)/(sin^2x-sin^4x)
=[(1-(sin^4x+cos^4x-sin²xcos²x)]/[sin²x(1-sin²x)]
=[1-(1-3sin²xcos²x)]/[sin²x(1-sin²x)]
=3sin²xcos²x/sin²xcos²x
=3

题目出错了