Question

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

Answer 1

Answer: 1

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