sin (A-B) sin (B-C) sin (C-A) = ?
Answers
Answered by
1
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) ]
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) ]
kvnmurty:
click on thanks azur blue button above please
Similar questions