求证sin²α+sin²β-sin²α*sin²β+cos²α*cos²β=1
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![求证sin²α+sin²β-sin²α*sin²β+cos²α*cos²β=1](/uploads/image/z/2739193-25-3.jpg?t=%E6%B1%82%E8%AF%81sin%26%23178%3B%CE%B1%2Bsin%26%23178%3B%CE%B2-sin%26%23178%3B%CE%B1%2Asin%26%23178%3B%CE%B2%2Bcos%26%23178%3B%CE%B1%2Acos%26%23178%3B%CE%B2%3D1)
求证sin²α+sin²β-sin²α*sin²β+cos²α*cos²β=1
求证sin²α+sin²β-sin²α*sin²β+cos²α*cos²β=1
求证sin²α+sin²β-sin²α*sin²β+cos²α*cos²β=1
证明:sin²α+sin²β-sin²α*sin²β+cos²α*cos²β
=sin²α(1-sin²β)+sin²β+cos²α*cos²β
=sin²α*cos²β+cos²α*cos²β+sin²β
=cos²β*(sin²α+cos²α)+sin²β
=cos²β+sin²β
=1
等式得证!