Question

Period of $\cos x·\cos (60°-x)·\cos (60°+x)$ is:

• A. $\frac{\mathrm{\pi }}{2}$
• B. $\mathrm{\pi }$
• C. $\frac{\mathrm{\pi }}{3}$
• D. $\frac{2\mathrm{\pi }}{3}$

Answer 1

The correct option is D

$\frac{2\mathrm{\pi }}{3}$

The explanation for the correct option.

Let,

$$\begin{gathered} f\left(x\right)=\cos x·\cos (60°-x)·\cos (60°+x) \\ \Rightarrow =\cos x\left(\frac{\cos 120°+\cos 2x}{2}\right)\mathbf{}\mathbf{}\left[\because 2cosAcosB=cos(A+B)+cos(A-B)\right] \\ \Rightarrow =\cos x\left(\frac{\cos 2x-{\displaystyle \frac{1}{2}}}{2}\right)\left[\because cos120°=-\frac{1}{2}\right] \\ \Rightarrow =\cos x\left(\frac{2\cos 2x-1}{4}\right) \\ \Rightarrow =\cos x\left(\frac{2(2{\cos }^{2}x-1)-1}{4}\right)\left[\because cos2x=2co{s}^{2}x-1\right] \\ \Rightarrow =\cos x\left(\frac{4{\cos }^{2}x-3}{4}\right) \\ \Rightarrow =\left(\frac{4{\cos }^{3}x-3\cos x}{4}\right) \\ \Rightarrow =\frac{\cos 3x}{4}\left[\because cos3x=4co{s}^{3}x-3cosx\right] \end{gathered}$$

Compare $f\left(x\right)=\frac{\cos 3x}{4}$ from $\frac{\cos kx}{4}$, We get $k=3$.

$$\begin{gathered} \therefore \mathrm{Period}=\frac{2\mathrm{\pi }}{\left|k\right|} \\ =\frac{2\mathrm{\pi }}{3} \end{gathered}$$

Hence, the option $D$ is correct.