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April 23rd, 2011, 05:47 AM   #1
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prove the trigonometric identity

if sin a + sin b + sin c = cos a + cos b + cos c = 0 then prove the following:

cos 3a + cos 3b + cos 3c = 3cos(a+b+c)
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April 23rd, 2011, 08:32 AM   #2
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1 = sina + cosa
= (-sin(b) - sin(c)) + (-cos(b) - cos(c))
= sinb + 2sin(b)sin(c) + sinc + cosb + 2cos(b)cos(c) + cosc
= 2 + 2sin(b)sin(c) + 2cos(b)cos(c),
so 2sin(b)sin(c) = -1 - 2cos(b)cos(c).

Expand 3cos(a+b+c) and eliminate all references to sin(a), cos(a), sinb, sinc and sin(b)sin(c). What do you get?
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