Trigonometry Proof

Date: 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)


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