Question

求证：sin²A+sin²B-sin²A·sin²B+cos²A·cos²B=1

Answer 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

Answer 2

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

Answer 3

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