Search any question & find its solution
Question:
Answered & Verified by Expert
$\sqrt{2+\sqrt{2+\sqrt{2+2 \cos 8 \theta}}}=$
Options:
Solution:
1775 Upvotes
Verified Answer
The correct answer is:
$2 \cos \theta$
Let $y=\sqrt{2+\sqrt{2+\sqrt{2+2 \cos 8 \theta}}}$
We know, $1+\cos 2 A=2 \cos ^2 A$
Therefore, $y=\sqrt{2+\sqrt{2+\sqrt{2 \cdot(1+\cos 8 \theta)}}}$
$$
\begin{aligned}
& =\sqrt{2+\sqrt{2+\sqrt{2 \cdot 2 \cos ^2 4 \theta}}} \\
& =\sqrt{2+\sqrt{2+2 \cos 4 \theta}}=\sqrt{2+\sqrt{2 \cdot 2 \cos ^2 2 \theta}} \\
& =\sqrt{2+2 \cos 2 \theta}=\sqrt{2 \cdot 2 \cos ^2 \theta}=2 \cos \theta
\end{aligned}
$$
We know, $1+\cos 2 A=2 \cos ^2 A$
Therefore, $y=\sqrt{2+\sqrt{2+\sqrt{2 \cdot(1+\cos 8 \theta)}}}$
$$
\begin{aligned}
& =\sqrt{2+\sqrt{2+\sqrt{2 \cdot 2 \cos ^2 4 \theta}}} \\
& =\sqrt{2+\sqrt{2+2 \cos 4 \theta}}=\sqrt{2+\sqrt{2 \cdot 2 \cos ^2 2 \theta}} \\
& =\sqrt{2+2 \cos 2 \theta}=\sqrt{2 \cdot 2 \cos ^2 \theta}=2 \cos \theta
\end{aligned}
$$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.