sin²α·cos²β＋cos²α·cos²β-1/2cos2α·cos2β的值

来源：学生作业学帮网编辑：学帮网时间：2024/05/31 02:59:28

=cos²β(sin²α+cos²α)-1/2cos2α·cos2β
=1/2(1+cos2β)-1/2cos2α·cos2β
=1/2(1+2sin²αcos2β)

=sin 2;αsin 2;β+cos 2;αcos 2;β-2cos 2;αcos 2;β+cos 2;α+cos 2;β-1/2 =sin 2;αsin 2;β-cos 2;αcos 2;β+cos 2;α