# Sin (A-B) sin (B-C) sin (C-A) = ?

1
by khakhramm

2015-05-01T00:52:33+05:30

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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) ]

thanks. but is this the answer that is expected ? or any other expression is expected...?
i am sure that u will be more helpful to me for my inter subjects too :)
i think soo this is the answer which the user may expected sir
ok.
but i am not sure coz it is not belongs to my standard :P