Question

What is the period of the curve whose equation is $y=\frac{1}{3}({\cos }^{2}x-{\sin }^{2}x)$?

• A. ${60}^o$
• B. ${120}^o$
• C. ${180}^o$
• D. ${360}^o$
• E. ${720}^o$

Answer 1

Answer: (C)

${180}^o$

Given, $y=\frac{1}{3}({\cos }^{2}x-{\sin }^{2}x)=\frac{1}{3}\cos 2x$
The period of $\cos x$ is $2\pi$
Therefore the period of $\cos 2x$ is $\frac{2\pi }{2}=\pi$
Therefore the period of $y$ is $\pi =180$.