问题描述: 求证sin²a+sin²b-sin²asin²b+cos²acos²b=1 1个回答 分类:数学 2014-10-31 问题解答: 我来补答 证明:sin^2A+sin^2B-sin^2Asin^2B+cos^2Acos^2B=sin^2A+sin^2B+cos^2Acos^2B-sin^2Asin^2B=sin^2A+sin^2B+(cosAcos2B-sinAsinB)(cosAcos2B+sinAsinB)=sin^2A+sin^2B+cos(A+B)cos(A-B)=(1-cos2A)/2+(1-cos2B)/2+cos(A+B)cos(A-B)=-(cos2A+cos2B)/2+cos(A+B)cos(A-B)+1=-2 cos[(2A+2B)/2] cos[(2A-2B)/2] /2+cos(A+B)cos(A-B)+1=-cos(A+B)cos(A-B)+cos(A+B)cos(A-B)+1=1得证希望我的回答对你有帮助,采纳吧O(∩_∩)O! 展开全文阅读