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*Are these angles A, B and C angles of a triangle?*we know Sin A Sin B = [ Cos (A-B) - Cos (A+B) ] / 2

and, Sin A Cos B = [ Sin (A+B) + Sin (A-B) ] / 2

Sin (A - B) Sin (B - C) = [ Cos (A+C - 2B) - Cos (A - C) ] / 2

Sin (A - B) Sin (B - C) Sin (C - A)

= [ Cos (A+C-2B) - Cos (A-C) ] / 2 * Sin (C - A) expand the terms

= 1/2 * Sin (C-A) Cos (A+C-2B) - 1/2 * Sin (C-A) Cos (A-C)

= 1/4 * [ Sin (C-A+A+C-2B) + Sin (C-A-A-C+2B) ] - 1/4 Sin (2C-2A)

*= 1/4 * [ Sin 2(C-B) + Sin 2(B-A) - Sin 2(C-A) ]*if A+B+C = 180 deg, and they are angles in a triangle, then,

= 1/4 * [ Sin (360 - 2A - 4B) + Sin (2B-2A) - Sin (360 - 4A - 2B) ]

*= 1/4 * [ - Sin (2A+4B) + Sin (2B - 2A) + Sin (4A+ 2B) ]*