Question

In ∆ABC, prove the following:
${\sin }^{3}A\cos \left(B-C\right)+{\sin }^{3}B\cos \left(C-A\right)+{\sin }^{3}C\cos \left(A-B\right)=3\sin A\sin B\sin C$

Answer 1

$$\begin{gathered} \mathrm{Let}\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=k...\left(1\right) \\ \mathrm{LHS}={\sin }^{3}A\cos \left(B-C\right)+{\sin }^{3}B\cos \left(C-A\right)+{\sin }^{3}C\cos \left(A-B\right) \\ ={\sin }^{2}A\left\{\sin A\cos \left(B-C\right)\right\}+{\sin }^{2}B\left\{\sin B\cos \left(C-A\right)\right\}+{\sin }^{2}A\left\{\sin A\cos \left(A-B\right)\right\} \\ =\frac{a^2}{k^2}\left\{\sin A\cos \left(B-C\right)\right\}+\frac{b^2}{k^2}\left\{\sin B\cos \left(C-A\right)\right\}+\frac{c^2}{k^2}\left\{\sin A\cos \left(A-B\right)\right\}\left[\mathrm{from}\left(1\right)\right] \\ =\frac{1}{2k^2}\left[a^2\left\{2\sin A\cos \left(B-C\right)\right\}+b^2\left\{2\sin B\cos \left(C-A\right)\right\}+c^2\left\{2\sin A\cos \left(A-B\right)\right\}\right] \\ =\frac{1}{2k^2}\left[a^2\left\{2\sin \left(\mathrm{\pi }-\left(B\mathit{+}C\right)\right)\cos \left(B-C\right)\right\}+b^2\left\{2\sin \left(\mathrm{\pi }-\left(A+C\right)\right)\cos \left(C-A\right)\right\}+c^2\left\{2\sin \left(\mathrm{\pi }-\left(B+C\right)\right)\cos \left(A-B\right)\right\}\right] \\ =\frac{1}{2k^2}\left[a^2\left\{2\sin \left(B+C\right)\cos \left(B-C\right)\right\}+b^2\left\{2\sin \left(C+A\right)\cos \left(C-A\right)\right\}+c^2\left\{2\sin \left(A+B\right)\cos \left(A-B\right)\right\}\right] \\ =\frac{1}{2k^2}\left[a^2\left\{\sin 2B+\sin 2C\right\}+b^2\left\{\sin 2C+\sin 2A\right\}+c^2\left\{\sin 2A+\sin 2B\right\}\right] \\ =\frac{1}{2k^2}\left[2a^2\left\{\sin B\cos B+\sin C\cos C\right\}+2b^2\left\{\sin C\cos C+\sin A\cos A\right\}+2c^2\left\{\sin A\cos A+\sin B\cos B\right\}\right] \\ =\frac{1}{2k^3}\left[2a^2\left\{k\sin B\cos B+k\sin C\cos C\right\}+2b^2\left\{k\sin C\cos C+k\sin A\cos A\right\}+2c^2\left\{k\sin A\cos A+k\sin B\cos B\right\}\right] \\ =\frac{1}{k^3}\left[a^2\left\{b\cos B+c\cos C\right\}+2b^2\left\{c\cos C+a\cos A\right\}+2c^2\left\{a\cos A+a\cos B\right\}\right] \\ =\frac{1}{k^3}\left[ab\left(a\cos B+b\cos A\right)+bc\left(b\cos C+c\cos B\right)+ac\left(a\cos C+c\cos A\right)\right] \\ =\frac{1}{k^3}\left(abc+bca+acb\right) \\ =3abc\times \frac{1}{k^3} \\ =3\sin A\sin B\sin C\times \frac{1}{k^3}\times k^3 \\ =3\sin A\sin B\sin C \\ =\mathrm{RHS} \\ \mathrm{Hence}\mathrm{proved}. \end{gathered}$$