Cosines everywhere (Posted on 2005-03-21)

	Submitted by Federico Kereki
Rating: 4.5000 (2 votes)
Solution:	(Hide)
We have 1-8.cos(A).cos(B).cos(C) = 1-4.cos(A)(cos(B-C)+cos(B+C)). Since cos(B+C)= -cos(π-B-C)= -cos(A), the expression above is: 1-4.cos(A).cos(B-C) + 4.cos(A)² = sin(B-C)² + cos(B-C)² -4.cos(A).cos(B-C) + 4.cos(A)² = sin(B-C)² + [cos(B-C)-2cos(A)]² which is always non negative.

	Subject	Author	Date
	probable mistake	John	2006-10-16 01:18:32
	re(2): This problem needs some formatting changes...	Erik O.	2005-04-05 22:38:46
	Advanced Calculus Problem	Richard	2005-03-23 22:55:31
	re: This problem needs some formatting changes...	Richard	2005-03-23 03:07:33
	This problem needs some formatting changes...	Erik O.	2005-03-22 17:25:46
	Solution	David Shin	2005-03-21 19:13:57