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April 23rd, 2011, 05:47 AM  #1 
Newbie Joined: Apr 2011 Posts: 1 Thanks: 0  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) 
April 23rd, 2011, 08:32 AM  #2 
Global Moderator Joined: Dec 2006 Posts: 19,532 Thanks: 1750 
1 = sin²a + cos²a = (sin(b)  sin(c))² + (cos(b)  cos(c))² = sin²b + 2sin(b)sin(c) + sin²c + cos²b + 2cos(b)cos(c) + cos²c = 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), sin²b, sin²c and sin(b)sin(c). What do you get? 

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identity, prove, trigonometric 
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