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.
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.
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 $a$,$b$,$c$.