Question

求证sin²a+sin²b-sin²asin²b+cos²acos²b=1

Answer 1

证明：
sin^2A+sin^2B-sin^2Asin^2B+cos^2Acos^2B
=sin^2A+sin^2B+cos^2Acos^2B-sin^2Asin^2B
=sin^2A+sin^2B+(cosAcos2B-sinAsinB)(cosAcos2B+sinAsinB)
=sin^2A+sin^2B+cos(A+B)cos(A-B)
=(1-cos2A)/2+(1-cos2B)/2+cos(A+B)cos(A-B)
=-(cos2A+cos2B)/2+cos(A+B)cos(A-B)+1
=-2 cos[(2A+2B)/2] cos[(2A-2B)/2] /2+cos(A+B)cos(A-B)+1
=-cos(A+B)cos(A-B)+cos(A+B)cos(A-B)+1
=1
得证

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