Question

The period of the function $f\left(x\right)=\frac{[\sin \left(8x\right)\cos \left(x\right)-\sin \left(6x\right)\cos \left(3x\right)]}{[\cos \left(2x\right)\cos \left(x\right)-\sin \left(3x\right)\sin \left(4x\right)]}$

• A. $\mathrm{\pi }$
• B. $2\mathrm{\pi }$
• C. $\frac{\mathrm{\pi }}{2}$
• D. None of these

Answer 1

The correct option is C

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

Explanation for the correct option:

Compute the period of the given function.

$f\left(x\right)=\frac{[\sin \left(8x\right)\cos \left(x\right)-\sin \left(6x\right)\cos \left(3x\right)]}{[\cos \left(2x\right)\cos \left(x\right)-\sin \left(3x\right)\sin \left(4x\right)]}$

$=\frac{[\sin \left(9x\right)+\sin \left(7x\right)]–[\sin \left(9x\right)+\sin \left(3x\right)]}{[\cos \left(3x\right)+\cos \left(x\right)]+[\cos \left(7x\right)–\cos \left(x\right)]}$

$=\frac{[\sin \left(7x\right)–\sin \left(3x\right)]}{[\cos \left(7x\right)+\cos \left(3x\right)]}$

$$\begin{gathered} =\frac{2\cos \left(5x\right)\sin \left(2x\right)}{2\cos \left(2x\right)\cos \left(5x\right)} \\ =\tan \left(2x\right) \end{gathered}$$

The general formula for the period of $\tan \left(ax\right)$ is $\frac{\mathrm{\pi }}{a}$.

Then the period of $tan\left(2x\right)$ is $\frac{\mathrm{\pi }}{2}$ .

Thus, the period of $f\left(x\right)$ is $\frac{\mathrm{\pi }}{2}$.

Hence, Option $\left(\mathrm{C}\right)$ is the correct answer.