Question

If $\cos A , \cos B$ and $\cos C$ are the roots of the cubic $x ^3+ ax ^2+ bx + c =0$ where $A , B , C$ are the angles of a triangle then find the value of $a^2-2 b-2 c$.

Answer 1

$ \sum \cos A =- a ; \sum \cos A \cos B = b \text { and } \prod \cos A =- c $
$\text { now } (\cos A +\cos B +\cos C )^2=\left(\sum \cos ^2 A \right)+2\left(\sum \cos A \cos B \right) $
$\therefore \cos ^2 A +\cos ^2 B +\cos ^2 C = a ^2-2 b$
$1-2 \cos A \cos B \cos C = a ^2-2 b ^2 $
$1-2 c = a ^2-2 b ^2 \Rightarrow a ^2-2 b -2 c =1 $