Q. For any angle $\theta$, the expression $\frac{2\cos{8\theta + 1}}{2\cos{\theta + 1}}$ is equal to
A) $\left( 2\cos{\theta + 1} \right)\left( 2\cos{2\theta + 1} \right)(2\cos{4\theta + 1})$
B) $\left( \cos{\theta - 1} \right)\left( \cos{2\theta - 1} \right)(\cos{4\theta - 1})$
C) $\left( 2\cos{\theta - 1} \right)\left( 2\cos{2\theta - 1} \right)(2\cos{4\theta - 1})$
D) $\left( 2\cos{\theta + 1} \right)\left( 2\cos{2\theta + 1} \right)(2\cos{4\theta + 1})$
Explanation
(c)
$\frac{2\cos{8\theta + 1}}{2\cos{\theta + 1}} = \frac{2\left( 2\cos^{2}{4\theta - 1} \right) + 1}{(2\cos{\theta + 1})}$
$= \frac{\left( 2\cos{4\theta - 1} \right)(2\cos{4\theta + 1})}{(2\cos{\theta + 1})}$
$= \frac{\left( 2\cos{4\theta - 1} \right)\left( 2\cos{2\theta - 1} \right)(2\cos{2\theta + 1})}{(2\cos{\theta + 1})}$
$= \frac{\left\lbrack \left( 2\cos{4\theta - 1} \right)\left( 2\cos{2\theta - 1} \right)\left( 2\cos{\theta - 1} \right)(2\cos{\theta + 1}) \right\rbrack}{(2\cos{\theta + 1})}$
$= \left( 2\cos{4\theta - 1} \right)\left( 2\cos{2\theta - 1} \right)(2\cos{\theta - 1})$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.