Answer:
The identity sinA sin(A+2B) - sinB sin(B+2A) = (sinA)^2 - (sinB)^2 is proved below.
Explanation:
sinA sin(A+2B) - sinB sin(B+2A) =
sinA ( sin(A) cos(2B) + sin(2B) cos(A) ) - sinB (sin(B)cos(2A) + sin(2A)cos(B) ) =
sinA( sinA [cos^2(B)- sin^2(B)] + 2 sin(B) cos(B) cosA ) - sinB( sinB(cos^2A-sin^2A)+2 sinA cosA cos(B) ) =
sin^2A(cos^2B-sin^2B) + 2 sinB cosB sin(A) cos(A) - sin^2B(cos^2A-sin^2A) - 2 sinA sinB cosA cosB =
sin^2A(cos^2B-sin^2B) - sin^2B(cos^2A-sin^2A) =
sin^2Acos^2B - sin^2Asin^2B - sin^2Bcos^2A + sin^2Bsin^2A =
sin^2Acos^2B - sin^2Bcos^2A =
(1 - cos^2A)(cos^2B) - (1 - cos^2B)(cos^2A) =
cos^2B - cos^2A cos^2B - cos^2A + cos^2B cos^2A =
cos^2B - cos^2A =
1 - sin^2B - (1 - sin^2A) =
sin^2 (A) - sin^2 (B)