求证sin²α+sin²β-sin²α*sin²β+cos²α*cos²β=1
网友回答
sin²α+sin²β-sin²α*sin²β+cos²α*cos²β
=sin²α-sin²α*sin²β+sin²β+cos²α*cos²β
=sin²α(1-sin²β)+cos²α*cos²β+sin²β
=sin²α*cos²β+cos²α*cos²β+sin²β
=cos²β(sin²α+cos²α)+sin²β
=cos²β+sin²β
=1证毕.======以下答案可供参考======
供参考答案1:
sin²α+sin²β-sin²α*sin²β+cos²α*cos²β
=sin²α（1-sin²β）+sin²β+cos²α*cos²β
=sin²α*cos²β+sin²β+cos²α*cos²β
=cos²β*（sin²α+cos²α）+sin²β
=cos²β+sin²β
=1供参考答案2:
证明：sin²α+sin²β-sin²α*sin²β+cos²α*cos²β
=sin²α-sin²α*sin²β+cos²α*cos²β+sin²β
=sin²α*(1-sin²β)+cos²α*cos²β+sin²β
=sin²α*cos²β+cos²α*cos²β+sin²β
=(sin²α+cos²α)*cos²β+sin²β
=cos²β+sin²β=1