Q. If$\ 0{^\circ} < \theta < 180{^\circ},\ \ then\sqrt{2 + \sqrt{2 + \sqrt{2 + ... + \sqrt{2\left( 1 + \cos\theta \right),}}}}$ then being $n$ number of 2’s, is equal to
A) $2\cos\left( \frac{\theta}{2^{n}} \right)$
B) $2\cos\left( \frac{\theta}{2^{n - 1}} \right)$
C) $2\cos\left( \frac{\theta}{2^{n + 1}} \right)$
D) None of these
Explanation
(a)
$\sqrt{2 + \sqrt{2 + \sqrt{2 + \ldots + \sqrt{2(1 + \cos{\theta)}}}}}$ ($n\$numbers of 2”s)
$= \sqrt{2 + \sqrt{2 + \sqrt{2 + ... + \sqrt{\left( 2 + 2\cos\frac{\theta}{2} \right)}}}}\ \ \ \left\lbrack (n - 1)numbers\ of\ 2's \right\rbrack$
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$= \sqrt{2 + 2\cos{(\theta/2^{n - 1})\ }}$
$= \sqrt{2\left\{ 1 + 2\cos^{2}{(\theta/2^{n})} - 1 \right\}} = 2\cos{(\theta/2^{n})}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.