For any triangle, derive the formula $a=b\cos C+c\cos B$

$\left[Hint:Usethefactthat\sin A=\sin \left({180}^{\circ }-B-C\right)\right]$

$$Use $\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}LawofSines$

Sum of angles in a triangle $A+B+C={180}^{\circ }$

Sum formula $\sin \left(x+y\right)=\sin x\cos y+\cos x\sin y$

Co-function Identity $$\begin{gathered} \cos \left({90}^{\circ }-\theta \right)=\sin \theta \\ \sin \left({90}^{\circ }-\theta \right)=\cos \theta \end{gathered}$$

$$\begin{gathered} \frac{\sin A}{a}=\frac{\sin B}{b} \\ \frac{a}{\sin A}=\frac{b}{\sin B} \\ a=\frac{b·\sin A}{\sin B} \end{gathered}$$

$$\begin{gathered} \frac{\sin A}{a}=\frac{\sin C}{c} \\ \frac{a}{\sin A}=\frac{c}{\sin C} \\ a=\frac{c·\sin A}{\sin C} \end{gathered}$$

$$\begin{gathered} a+a=\frac{b·\sin A}{\sin B}+\frac{c·\sin A}{\sin C} \\ 2a=\frac{b·\sin A}{\sin B}+\frac{c·\sin A}{\sin C} \end{gathered}$$

$$\begin{gathered} \sin A=\sin \left({180}^{\circ }-B-C\right) \\ \sin A=\sin \left({90}^{\circ }+{90}^{\circ }-B-C\right) \\ \sin A=\sin \left[\left({90}^{\circ }-B\right)+\left({90}^{\circ }-C\right)\right] \end{gathered}$$

Use the sum formula

$$\begin{gathered} \sin A=\sin \left[\left({90}^{\circ }-B\right)+\left({90}^{\circ }-C\right)\right] \\ \sin A=\sin \left({90}^{\circ }-B\right)\cos \left({90}^{\circ }-C\right)+\cos \left({90}^{\circ }-B\right)\sin \left({90}^{\circ }-C\right) \\ \sin A=\cos B\sin C+\sin B\cos C \end{gathered}$$

$2a=\frac{b·\sin A}{\sin B}+\frac{c·\sin A}{\sin C}$

$$\begin{gathered} 2a=\frac{b\left(\cos B\sin C+\sin B\cos C\right)}{\sin B}+\frac{c\left(\cos B\sin C+\sin B\cos C\right)}{\sin B} \\ 2a=\frac{\left(b\cos B\sin C+b\sin B\cos C\right)}{\sin B}+\frac{\left(c\cos B\sin C+c\sin B\cos C\right)}{\sin B} \end{gathered}$$

$$\begin{gathered} 2a=\frac{b\cos B\sin C}{\sin B}+\frac{b\sin B\cos C}{\sin B}+\frac{c\cos B\sin C}{\sin C}+\frac{c\sin B\cos C}{\sin C} \\ 2a=\frac{b\cos B\sin C}{\sin B}+b\cos C+c\cos B+\frac{c\sin B\cos C}{\sin C}----\left(1\right) \end{gathered}$$

$$\begin{gathered} \frac{b\cos B\sin C}{\sin B}=\frac{c\cos B\sin C}{\sin C}\left[SInce\frac{b}{\sin B}=\frac{c}{\sin C}\right] \\ =c\cos B \end{gathered}$$

$\frac{c\sin B\cos C}{\sin C}=\frac{b\sin B\cos C}{\sin B}=b\cos C$

$$\begin{gathered} 2a=c\cos B+b\cos C+c\cos B+b\cos C \\ 2a=2b\cos C+2c\cos B \\ a=b\cos C+c\cos B \end{gathered}$$