Question

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

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ \left|\mathrm{cos\theta }\cos \left(60°-\mathrm{\theta }\right)\cos \left(60°+\mathrm{\theta }\right)\right| \\ =\left|\mathrm{cos\theta }\left({\cos }^{2}60°-{\sin }^2\mathrm{\theta }\right)\right|\left[{\cos }^2\mathrm{A}-{\sin }^2\mathrm{B}=\cos \left(\mathrm{A}-\mathrm{B}\right)\cos \left(\mathrm{A}+\mathrm{B}\right)\right] \\ =\left|\mathrm{cos\theta }\left(\frac{1}{4}-{\sin }^2\mathrm{\theta }\right)\right| \\ =\left|\mathrm{cos\theta }\frac{1}{4}\left(1-4{\sin }^2\mathrm{\theta }\right)\right| \\ =\left|\frac{1}{4}\mathrm{cos\theta }\left\{1-4\left(1-{\cos }^2\mathrm{\theta }\right)\right\}\right| \\ =\left|\frac{1}{4}\mathrm{cos\theta }\left\{-3+4{\cos }^2\mathrm{\theta }\right\}\right| \end{gathered}$$

$$\begin{gathered} =\left|\frac{1}{4}\left(4{\cos }^3\mathrm{\theta }-3\mathrm{cos\theta }\right)\right| \\ =\left|\frac{1}{4}\cos 3\mathrm{\theta }\right|\left[\because \cos 3\mathrm{\theta }=4{\cos }^3\mathrm{\theta }-3\mathrm{cos\theta }\right] \\ \le \frac{1}{4}\left(\because \left|\cos x\right|\le 1\mathrm{for}\mathrm{all}x\right) \\ \therefore \left|\mathrm{cos\theta }\cos \left(60°-\mathrm{\theta }\right)\cos \left(60°+\mathrm{\theta }\right)\right|\le \frac{1}{4} \end{gathered}$$