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 Linear Algebra Linear Algebra Math Forum

 September 12th, 2018, 07:33 AM #1 Newbie   Joined: Jul 2018 From: Asia Posts: 9 Thanks: 1 simultaneous equations involving cosine Could anyone help with solving these two simultaneous equations to obtain $\displaystyle k=2\sin(B/2)$ ? $\displaystyle \cos(AB) - \cos[(A-1)B] + k = 0$ $\displaystyle B = π(2k-1)/(2A-1)$ Last edited by skipjack; September 13th, 2018 at 02:06 AM. September 13th, 2018, 03:12 AM #2 Global Moderator   Joined: Dec 2006 Posts: 20,384 Thanks: 2011 If you've given the problem correctly and in full, it can't be proved that $k = 2\sin(B/2)$. If $A = -1$, $B =-\pi$ and $k = 2$, the simultaneous equations are satisfied, but $2\sin(B/2) = -2$, not $2$. The same applies if $A = 1$, $B =3\pi$ and $k = 2$. September 13th, 2018, 07:47 AM #3 Newbie   Joined: Jul 2018 From: Asia Posts: 9 Thanks: 1 The full derivation solution can be found at https://www.reddit.com/r/learnmath/c...olving_cosine/  Tags cosine, equations, involving, simultaneous Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post sothirtyyearsago Calculus 6 February 25th, 2017 09:06 PM jamesbrown Algebra 1 January 6th, 2015 08:37 AM srahman33 Algebra 2 August 28th, 2014 02:55 AM queenie_n Complex Analysis 5 October 17th, 2012 05:20 PM empiricus Algebra 4 August 30th, 2009 04:28 AM

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