Question 570773
A,B,C can form a triangle so use law of cosines, where a,b,c are the sides opposite A,B,C respectively.


*[tex \LARGE a^2 = b^2 + c^2 - 2bc \cos A \Rightarrow \cos A = \frac{a^2 - b^2 - c^2}{2bc}]


Similarly,


*[tex \LARGE \cos B = \frac{b^2 - a^2 - c^2}{2ac}]


*[tex \LARGE \cos C = \frac{c^2 - a^2 - b^2}{2ab}]


Replace this into your equation and you (should) obtain an identity. However, this part is awfully long so a more elegant solution would be nice. A similar solution would be to use the law of sines:


*[tex \LARGE \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R] where R is the circumradius. Therefore


*[tex \LARGE \sin A = \frac{a}{2R} \Rightarrow \cos A = \pm \sqrt{1 - \sin^2 A} = \pm \sqrt{1 - \frac{a^2}{4R^2}} \Rightarrow \cos^2 A = 1 - \frac{a^2}{4R^2}]. Also another long solution.