2006-05-10T08:42:06ZFluxBBhttp://www.mathisfunforum.com/viewtopic.php?id=3546maybe this will help]]>http://www.mathisfunforum.com/profile.php?id=24462006-05-10T08:42:06Zhttp://www.mathisfunforum.com/viewtopic.php?pid=34570#p34570RHS=sin(A+B)=sinAcosB+cosAsinB If A+B=90, B=90-A, A=90-B; cosB=sin(90-B)=sinA sinB=cos(90-B)=cosA Therefore RHS=sin²A+cos²A=1 LHS=sin²A+sin²B, since sinB=cosA, sin²B=cos²A, therefore, LHS=1=RHS. We find the equation is true when A+B=90 degrees or (A+B)/2=180. Shall try to prove later....]]>http://www.mathisfunforum.com/profile.php?id=6822006-05-09T14:25:18Zhttp://www.mathisfunforum.com/viewtopic.php?pid=34505#p34505A and B are two acute angles. Prove that if sin² (A) + sin² (B) = sin( A + B ), then A + B = 180/2.