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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[(A1)B] + k = 0 $ $\displaystyle B = π(2k1)/(2A1) $ 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  

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