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September 12th, 2018, 08:33 AM   #1
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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 03:06 AM.
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September 13th, 2018, 04:12 AM   #2
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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 $k = 2$.
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September 13th, 2018, 08:47 AM   #3
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The full derivation solution can be found at https://www.reddit.com/r/learnmath/c...olving_cosine/

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