Question

${\sin }^{3}A+{\sin }^3\left(\frac{2\pi }{3}+A\right){\sin }^3\left(\frac{4\pi }{3}+A\right)=-\frac{3}{4}\sin 3A$

Answer 1

$$\begin{gathered} \mathrm{LHS}={\sin }^3\mathrm{A}+{\sin }^3\left(\frac{2\mathrm{\pi }}{3}+\mathrm{A}\right)+{\sin }^3\left(\frac{4\mathrm{\pi }}{3}+\mathrm{A}\right) \\ =\frac{3\mathrm{sinA}-\sin 3\mathrm{A}}{4}+\frac{3\sin \left({\displaystyle \frac{2\mathrm{\pi }}{3}}+\mathrm{A}\right)-\sin 3\left(\frac{2\mathrm{\pi }}{3}+\mathrm{A}\right)}{4}+\frac{3\sin \left({\displaystyle \frac{4\mathrm{\pi }}{3}}+\mathrm{A}\right)-\sin 3\left(\frac{4\mathrm{\pi }}{3}+\mathrm{A}\right)}{4} \\ \left[{\sin }^3\mathrm{\theta }=\frac{3\mathrm{sin\theta }-\sin 3\mathrm{\theta }}{4}\right] \\ =\frac{3\mathrm{sinA}-\sin 3\mathrm{A}}{4}+\frac{3\sin \left\{\mathrm{\pi }-\left(\frac{2\mathrm{\pi }}{3}+\mathrm{A}\right)\right\}-\sin \left(2\mathrm{\pi }+3\mathrm{A}\right)}{4}+\frac{3\sin \left\{\mathrm{\pi }+\left(\frac{\mathrm{\pi }}{3}+\mathrm{A}\right)\right\}-\sin \left(4\mathrm{\pi }+3\mathrm{A}\right)}{4} \\ =\frac{1}{4}\left[\left(3\mathrm{sinA}-\sin 3\mathrm{A}\right)+\left\{3\sin \left(\frac{\mathrm{\pi }}{3}-\mathrm{A}\right)-\sin 3\mathrm{A}\right\}-\left\{3\sin \left(\frac{\mathrm{\pi }}{3}+\mathrm{A}\right)+\sin 3\mathrm{A}\right\}\right] \\ =\frac{1}{4}\left[3\mathrm{sinA}-\sin 3\mathrm{A}+3\sin \left(\frac{\mathrm{\pi }}{3}-\mathrm{A}\right)-3\sin \left(\frac{\mathrm{\pi }}{3}+\mathrm{A}\right)-\sin 3\mathrm{A}-\sin 3\mathrm{A}\right] \end{gathered}$$

$$\begin{gathered} =\frac{1}{4}\left[3\mathrm{sinA}-3\sin 3\mathrm{A}+3\left\{\sin \left(\frac{\mathrm{\pi }}{3}-\mathrm{A}\right)-\sin \left(\frac{\mathrm{\pi }}{3}+\mathrm{A}\right)\right\}\right] \\ =\frac{1}{4}\left[3\mathrm{sinA}-3\sin 3\mathrm{A}+3\left\{2\cos \frac{{\displaystyle \frac{\mathrm{\pi }}{3}}-\mathrm{A}+{\displaystyle \frac{\mathrm{\pi }}{3}}+\mathrm{A}}{2}\sin \frac{{\displaystyle \frac{\mathrm{\pi }}{3}}-\mathrm{A}-{\displaystyle \frac{\mathrm{\pi }}{3}}-\mathrm{A}}{2}\right\}\right] \\ \left[\because \mathrm{sinC}-\mathrm{sinD}=2\cos \frac{\mathrm{C}+\mathrm{D}}{2}\sin \frac{\mathrm{C}-\mathrm{D}}{2}\right] \\ =\frac{1}{4}\left[3\mathrm{sinA}-3\sin 3\mathrm{A}+6\cos \frac{\mathrm{\pi }}{3}\sin \left(-\mathrm{A}\right)\right] \\ =\frac{1}{4}\left[3\mathrm{sinA}-3\sin 3\mathrm{A}-3\mathrm{sinA}\right] \\ =-\frac{3}{4}\mathrm{sinA} \\ =\mathrm{RHS} \\ \mathrm{Hence}\mathrm{proved}. \end{gathered}$$