My Math Forum prove the trigonometric identity
 User Name Remember Me? Password

 Trigonometry Trigonometry Math Forum

 April 23rd, 2011, 06:47 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, 09:32 AM #2 Global Moderator   Joined: Dec 2006 Posts: 20,310 Thanks: 1981 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?

 Tags identity, prove, trigonometric

 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post OriaG Trigonometry 10 August 12th, 2012 05:37 PM afrim Trigonometry 8 October 1st, 2011 02:59 AM chnixi Algebra 1 August 19th, 2009 04:52 AM munkiemunkie007 Algebra 2 May 4th, 2009 05:50 PM TG Algebra 2 March 7th, 2008 04:22 PM

 Contact - Home - Forums - Cryptocurrency Forum - Top