Question

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

Answer 1

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

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