Question

I : Period of $\sin \frac{\theta }{3}+\cos \frac{\theta }{2}$ is $12\pi$
II : Period of ${\cos }^{3}x+{\cos }^3({120}^0-x)+{\cos }^3({120}^0+x)$ is $\frac{\pi }{3}$

• A. Only I is true
• B. Both I and II are true
• C. Only II is true
• D. Neither I nor II is true

Answer 1

The correct option is A Only I is true

$\text{CASE I}$

$\sin \frac{\theta }{3}+\cos \frac{\theta }{2}$

Period of $\sin \frac{\theta }{3}=\frac{2\pi }{\frac{1}{3}}=6\pi$

Period of $\cos \frac{\theta }{2}=\frac{2\pi }{\frac{1}{2}}=4\pi$

$\therefore L.C.M.of6\pi and4\pi is12\pi$

$\therefore periodis12\pi$

$\text{verification:-}⟹\sin (12\pi +\frac{\theta }{3})+\cos (12\pi +\frac{\theta }{2})=\sin \frac{\theta }{3}+\cos \frac{\theta }{2}[\because \sin (2n\pi +\theta )=\sin \theta ,\cos (2n\pi +\theta )=\cos \theta ]$

$\text{CASE II}$

${\cos }^{3}x+{\cos }^3({120}^{\circ }-x)+{\cos }^3({120}^{\circ }+x)$

now, if we consider $\frac{\pi }{3}$ as period of above function, then

$⟹{\cos }^3(\frac{\pi }{3}+x)+{\cos }^3({120}^{\circ }-\frac{\pi }{3}-x)+{\cos }^3({120}^o+\frac{\pi }{3}+x)$

$⟹{\cos }^3(\frac{\pi }{3}+x)+{\cos }^3(\frac{\pi }{3}-x)+{\cos }^3(\pi +x)$

$⟹{\cos }^3(\frac{\pi }{3}+x)+{\cos }^3(\frac{\pi }{3}-x)-{\cos }^3\left(x\right)$

$(\cos (\pi \pm \theta )=-\cos \theta )$

$⟹{\cos }^3(\frac{\pi }{3}+x)+{\cos }^3(\frac{\pi }{3}-x)-{\cos }^3\left(x\right)\ne {\cos }^{3}x+{\cos }^3({120}^{\circ }-x)+{\cos }^3({120}^{\circ }+x)$

So Only $I$ is true.