Question

lf $f\left(x\right)={\sin }^2(\frac{\pi }{8}+\frac{x}{2})-{\sin }^2(\frac{\pi }{8}-\frac{x}{2})$, then the period of $f$ is

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

Answer 1

The correct option is D $2\pi$

$f\left(x\right)={\sin }^2(\frac{\pi }{8}+\frac{\pi }{2})-{\sin }^2(\frac{\pi }{8}-\frac{\pi }{2})$

$=[\sin (\frac{\pi }{8}+\frac{\pi }{2})+\sin (\frac{\pi }{8}-\frac{\pi }{2})]\times [\sin (\frac{\pi }{8}+\frac{\pi }{2})-\sin (\frac{\pi }{8}-\frac{\pi }{2})]$

$=\left[2\sin \left\{\frac{(\frac{\pi }{8}+\frac{x}{2}+\frac{\pi }{8}-\frac{x}{2})}{2}\right\}\cos \left\{\frac{(\frac{\pi }{8}+\frac{x}{2}-\frac{\pi }{8}+\frac{x}{2})}{2}\right\}\right]$

$\Rightarrow 2\sin \frac{\pi }{8}\cos \frac{x}{2}\times 2\cos \frac{\pi }{8}\sin \frac{x}{2}$

$\Rightarrow 2\sin \frac{\pi }{8}\cos \frac{\pi }{8}.2\sin \frac{x}{2}\cos \frac{x}{2}$

$\Rightarrow \sin \left(2.\frac{\pi }{8}\right).\sin \left(2.\frac{x}{2}\right)$

$\Rightarrow \sin \frac{\pi }{4}.\sin x=\frac{1}{\sqrt{2}}\sin x$

$f\left(x\right)=\frac{1}{\sqrt{2}}\sin x$

$\therefore$Period $=2\pi$ [Period of $\sin x=\frac{2\pi }{1}=2\pi$]