Prove that in a triangle ABC,:

sin(A)sin(B)sin(C) + cos(A)cos(B) = 1

implies:

A = B = 45° and C = 90°.

1=cos(a)cos(b)+sin(a)sin(b)sin(c)<=

<= sin(a)sin()+cos(a)cos(b)=cos(a-b) so a=b.

a,b>0, so equalityh holds iff sin(c)=1. Hence, result.

Posted by jacob on 2004-02-26 16:21:36