Precalculus (6th Edition) Blitzer

If $A$, $B$ and $C$ are the measures of the angle of a triangle, and $a$, $b$ and $c$ are the lengths of the sides opposite to these angles, then the Law of Cosines states that ${{a}^{2}}={{b}^{2}}+{{c}^{2}}-2bc\cos A$.
Consider the triangles given below and use the basic trigonometric ratios to obtain: From the figure, we get $a=c\cos B+b\cos C$ Similarly we get \begin{align} & b=c\cos A+a\cos C \\ & c=a\cos B+b\cos A \\ \end{align} Therefore, \begin{align} & {{a}^{2}}=ac\cos B+ab\operatorname{cosC} \\ & {{b}^{2}}=bc\cos A+ab\cos C \\ & {{c}^{2}}=ac\cos B+bc\cos A \\ \end{align} So, \begin{align} & {{b}^{2}}+{{c}^{2}}-{{a}^{2}}=ab\cos C+bc\cos A+bc\cos A+ac\cos B-ac\cos B-ab\cos C \\ & =2bc\cos A \end{align} Hence, ${{a}^{2}}={{b}^{2}}+{{c}^{2}}-2bc\cos A$.