Question

Prove that $\left|\sin \mathrm{\theta }\sin \left(60-\mathrm{\theta }\right)\sin \left(60+\mathrm{\theta }\right)\right|\le \frac{1}{4}$ for all values of θ

Answer 1

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