证明等式恒成立 sin^a+sin^b-sin^asin^b+cos^acos^b=1

题目

证明等式恒成立 sin^a+sin^b-sin^asin^b+cos^acos^b=1
sin^a+sin^b-sin^asin^b+cos^acos^b=1
只需证sin^a(1-sin^b)+sin^b+cos^acos^b=1
只需证sin^acos^b+cos^acos^b+sin^b=1
这里到这里没有看懂

答案

要证sin^a+sin^b-sin^asin^b+cos^acos^b=1只需证sin^a(1-sin^b)+sin^b+cos^acos^b=1--(合并同类项）只需证sin^acos^b+cos^acos^b+sin^b=1－－（因为sin^b+cos^b=1,1-sin^b=cos^b)只需证cos^b(sin^a+cos^a)+sin^b=1－...