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How to prove that

$1+2(\cos a)(\cos b)(\cos c)-\cos^2 a-\cos^2 b-\cos^2 c=4 (\sin p)(\sin q) (\sin r)(\sin s)$,


$p=\frac{1}{2}(-a+b+c)$, $q=\frac{1}{2}(a-b+c)$, $r=\frac{1}{2}(a+b-c)$, $s=\frac{1}{2}(a+b+c)$.


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1 Answer 1

up vote 4 down vote accepted


$$ \begin{eqnarray} 2 (\sin p)(\sin q) &=& \cos(p-q) - \cos(p+q) \\ 2 (\sin r)(\sin s) &=& \cos(r-s) - \cos(r+s) \\ \end{eqnarray} $$

Then use $$ \begin{eqnarray} \cos(p-q) \cos(r-s) &=& \frac{1}{2}( \cos(p+s-q-r) + \cos(p+r - s-q)) \\ \cos(p+q) \cos(r-s) &=& \frac{1}{2}( \cos(p+s+q-r) + \cos(p+r - s+q)) \\ \cos(p-q) \cos(r+s) &=& \frac{1}{2}( \cos(p-s-q-r) + \cos(p+r +s-q)) \\ \cos(p+q) \cos(r+s) &=& \frac{1}{2}( \cos(p-s+q-r) + \cos(p+r +s + q)) \end{eqnarray} $$ Now use expressions for $p$,$q$,$r$,$s$ in terms of $a$,$b$,$c$.

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Heh, prosthaphaeresis. :) – J. M. Sep 6 '11 at 17:42

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