TITLE: Some trigonometric formula
QUESTION [5 upvotes]: How to prove that
$1+2(\cos a)(\cos b)(\cos c)-\cos^2 a-\cos^2 b-\cos^2 c=4 (\sin p)(\sin q) (\sin r)(\sin s)$, 
where
$p=\frac{1}{2}(-a+b+c)$, $q=\frac{1}{2}(a-b+c)$, $r=\frac{1}{2}(a+b-c)$, $s=\frac{1}{2}(a+b+c)$.
Thanks.

REPLY [4 votes]: Use 
$$
\begin{eqnarray}
 2 (\sin p)(\sin q) &=&  \cos(p-q) - \cos(p+q)  \\
 2 (\sin r)(\sin s) &=&  \cos(r-s) - \cos(r+s)  \\
\end{eqnarray}
$$
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 $a$,$b$,$c$.