求证:sin²a+sin²β-sin²a×sin²β+cos²a×cos²β=1x是乘以
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![求证:sin²a+sin²β-sin²a×sin²β+cos²a×cos²β=1x是乘以](/uploads/image/z/2739194-26-4.jpg?t=%E6%B1%82%E8%AF%81%EF%BC%9Asin%26%23178%3Ba%2Bsin%26%23178%3B%CE%B2-sin%26%23178%3Ba%C3%97sin%26%23178%3B%CE%B2%2Bcos%26%23178%3Ba%C3%97cos%26%23178%3B%CE%B2%3D1x%E6%98%AF%E4%B9%98%E4%BB%A5)
求证:sin²a+sin²β-sin²a×sin²β+cos²a×cos²β=1x是乘以
求证:sin²a+sin²β-sin²a×sin²β+cos²a×cos²β=1
x是乘以
求证:sin²a+sin²β-sin²a×sin²β+cos²a×cos²β=1x是乘以
sin²a+sin²β-sin²a×sin²β+cos²a×cos²β
=sin²a(1-sin²β)+sin²β+cos²axcos²β
=sin²acos²β+cos²acos²β+sin²β
=(sin²a+cos²a)cos²β+sin²β
=cos²β+sin²β
=1
sin²a+sin²β-sin²a×sin²β+cos²a×cos²β
= sin²a+sin²β-sin²a×sin²β+(1-sin²a)(1-sin²β)
= sin²a+sin²β-sin²a×sin²β + 1-sin²a-sin²β+sin²a×sin²β
= 1,得证