Question

Prove that $\left|\sin x\sin \left(\frac{\pi }{3}-x\right)\sin \left(\frac{\pi }{3}+x\right)\right|\le \frac{1}{4}$ for all values of x

Answer 1

$$\begin{gathered} \frac{\pi }{3}=60° \\ \mathrm{We}\mathrm{have}, \\ \left|\sin x\sin \left(60-x\right)\sin \left(60+x\right)\right| \\ =\left|\sin x\left({\sin }^{2}60-{\sin }^{2}x\right)\right| \\ \left[\because \sin \left(\mathrm{A}+\mathrm{B}\right)\sin \left(\mathrm{A}-\mathrm{B}\right)={\sin }^2\mathrm{A}-{\sin }^2\mathrm{B}\right] \\ =\left|\sin x\left(\frac{3}{4}-{\sin }^{2}x\right)\right| \\ =\left|\frac{1}{4}\sin x\left(3-4{\sin }^{2}x\right)\right| \\ =\left|\frac{1}{4}3\sin x-4{\sin }^{3}x\right| \\ =\frac{1}{4}\left|\sin 3\mathrm{x}\right|\left(\because 3\sin x-4{\sin }^{3}x=\sin 3x\right) \\ \le \frac{1}{4}\left(\because \left|\sin x\right|\le 1\mathrm{for}\mathrm{all}x\right) \\ \therefore \left|\sin x\sin \left(60-x\right)\sin \left(60+x\right)\right|\le \frac{1}{4} \\ \mathrm{Hence}\mathrm{proved}. \end{gathered}$$