Question

数学。化简sin²a+sinb²-sin²asin²b+cos²acos²b ，详细过程

Answer 1

sin²a + sin²b - sin²asin²b + cos²acos²b
= (sin²a - sin²asin²b) + sin²b + cos²acos²b
= sin²a(1 - sin²b) + sin²b + cos²acos²b
= sin²a cos²b + sin²b + cos²acos²b
= (sin²a cos²b + cos²acos²b) + sin²b
= cos²b(sin²a + cos²a ) + sin²b
= cos²b + sin²b
= 1

Answer 2

sin²a+sinb²-sin²asin²b+cos²acos²b
=sin²a-sin²asin²b+sin²b+cos²acos²b
=sin²a(1-sinb²)+sin²b+cos²acos²b
=sin²acos²b+sin²b+cos²acos²b
=sin²acos²b+cos²acos²b+sin²b
=cos²b(sin²a+cos²a)+sin²b
=cos²b+sin²b
=1

Answer 3

原式=sin2a+sinb2-sin2asin2b+cos2acos2b
=sin2a（1-sin2b）+cos2acos2b
=sin2acos2b+cos2acos2b
=cos2b（sin2a+cos2a）
=cos2b