Question

Period of $\cos x$ $\cos ({60}^0-x)\cos ({60}^0+x)$ is

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

Answer 1

The correct option is C $\frac{2\pi }{3}$

Let $f\left(x\right)=\cos x\cos ({60}^o-x)\cos ({60}^o+x)$
$=\cos x\left[\frac{\cos \left(120\right)+\cos 2x}{2}\right][2\cos A\cos B=\cos (A+B)+\cos (A-B\left)\right]$
$=\cos x\left[\frac{\cos 2x-\frac{1}{2}}{2}\right]$
$=\cos x\left[\frac{2\cos 2x-1}{4}\right]$
$=\cos x\left[\frac{4{\cos }^{2}x-3}{4}\right][\cos 2x=2{\cos }^{2}x-1]$
$=\frac{4{\cos }^{3}x-3\cos x}{4}$
$f\left(x\right)=\frac{\cos 3x}{4}=\frac{\cos kx}{4}$
Period of $f\left(x\right)=\frac{2\pi }{|k|}=\frac{2\pi }{3}$