作业帮求证：sin²A+sin²B-sin²A·sin²B+cos²A·cos²B=1
来源：学生作业帮助网 编辑：作业帮 时间：2024/07/17 22:02:23
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求证：sin²A+sin²B-sin²A·sin²B+cos²A·cos²B=1
求证：sin²A+sin²B-sin²A·sin²B+cos²A·cos²B=1
求证：sin²A+sin²B-sin²A·sin²B+cos²A·cos²B=1
sin²A+sin²B-sin²A·sin²B+cos²A·cos²B
=sin²A(1-sin²B)+sin²B+cos²A·cos²B
=sin²A·cos²B+cos²A·cos²B+sin²B
=(sin²A+cos²A)cos²B+sin²B
=cos²B+sin²B
=1
左=(1-cos2A)/2+(1-cos2B)/2-(1-cos2A)(1-cos2B)/4+(1+cos2A)(1+cos2B)/4
=1-(cos2A+cos2B)/2+（(1+cos2A)(1+cos2B)-(1-cos2A)(1-cos2B)）/4
=1-(cos2A+cos2B)/2+（2(cos2A+cos2B)）/4
=1=右
sin∧2A sin∧2B(1-sin∧2A) cos∧2Acos∧2B=sin∧2A cos∧2Asin∧2B cos∧Acos∧2B=sin∧2A cos∧2A(sin∧2B cos∧2B)=sin∧2A cos∧2A=1