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 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/

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