Trigonometry ProofDate: 6/20/96 at 21:35:49 From: Anonymous Subject: Trigonometry Proof Prove that if a,b,y>0 and a+b+y = pi, then sin(2a)+sin(2b)+sin(2y) = 4(sin a)(sin b)(sin y) Thanks, -Michael Date: 6/21/96 at 7:52:16 From: Doctor Anthony Subject: Re: Trigonometry Proof Use the addition formula on sin(2a) + sin(2b) to give 2sin(a+b)cos(a-b) and then sin(a+b) = sin(pi-y) = sin(y). So we have: Left hand side = 2sin(y)cos(a-b) + sin(2y) use double angle formula on sin(2y): = 2sin(y)cos(a-b) + 2sin(y)cos(y) = 2sin(y){cos(a-b) + cos(y)} = 2sin(y){cos(a-b) - cos(pi-y)} = 2sin(y){2sin[(1/2)(a-b+pi-y)]sin[(1/2)(pi-y-a+b)]} = 4sin(y)sin[(1/2)(a+pi-b-y)]sin[(1/2)(b+pi-a-y)] = 4sin(y)sin[(1/2)(2a)]sin[(1/2)(2b)] = 4sin(y)sin(a)sin(b) -Doctor Anthony, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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