Question

In triangle ABC, prove the following:
$a\cos A+b\cos B+c\cos C=2b\sin A\sin C=2c\sin A\sin B$

Answer 1

$$\begin{gathered} \text{Let}\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=k...\left(1\right) \\ \mathrm{Consider}\mathrm{the}\mathrm{LHS}\mathrm{of}\mathrm{the}\mathrm{equation}a\cos A+b\cos B+c\cos C. \\ a\cos A+b\cos B+c\cos C=k\left(\sin A\cos A+\sin B\cos B+\sin C\cos C\right) \\ =\frac{k}{2}\left(2sinAcosA+2sinAcosA+2sinCcosC\right) \\ =\frac{k}{2}\left(\sin 2A+\sin 2B+\sin 2C\right) \\ =\frac{k}{2}\left[2\sin \left(A+B\right)\cos \left(A-B\right)+2\sin C\cos C\right] \\ =\frac{k}{2}\left[2\sin \left(\mathrm{\pi }-C\right)\cos \left(A-B\right)+2\sin C\cos C\right] \\ =\frac{k}{2}\left[2\sin C\cos \left(A-B\right)+2\sin C\cos C\right] \\ =\frac{2k\sin C}{2}\left[\cos \left(A-B\right)+\cos C\right] \end{gathered}$$
$$\begin{gathered} =k\sin C\left[\cos \left(A-B\right)+\cos \left\{\mathrm{\pi }-\left(\mathrm{A}+\mathrm{B}\right)\right\}\right] \\ =k\sin C\left[\cos \left(A-B\right)-\cos \left(\mathrm{A}+\mathrm{B}\right)\right] \\ =k\sin C\left[2\sin A\mathrm{sinB}\right] \\ =2k\sin A\sin B\sin C...\left(1\right) \end{gathered}$$



$$\begin{gathered} \mathrm{Now}, \\ \text{on}\mathrm{putting}k\sin C=C\mathrm{in}\mathrm{equation}\left(1\right),\mathrm{we}\mathrm{get}: \\ 2c\sin A\sin B \\ \mathrm{and}\text{on}\mathrm{putting}k\sin B=b\mathrm{in}\mathrm{equation}\left(1\right),\mathrm{we}\mathrm{get}: \\ 2b\sin A\sin C \end{gathered}$$

So, from (1), we have
$a\cos A+b\cos B+c\cos C=2b\sin A\sin C=2c\sin A\sin B$.

Hence proved.