Question

If $A+B+C={180}^0$, then ${\cos }^{2}A+{\cos }^{2}B+{\cos }^{2}C=1-s\cos A\cos B\cos C$
Find $s$

Answer 1

${\cos }^{2}A+{\cos }^{2}B+{\cos }^{2}C=\frac{1+\cos 2A}{2}+\frac{1+\cos 2B}{2}+\frac{1+\cos 2C}{2}$

$=\frac{1}{2}(\cos 2A+\cos 2B+\cos 2C)+\frac{3}{2}$

$=\frac{1}{2}(-1-4\cos A\cos B\cos C)+\frac{3}{2}$

$=1-2\cos A\cos B\cos C$
Ans: 2