Q70 / 180 · KCET 2022 · Mathematics
$\sqrt{2+\sqrt{2+\sqrt{2+2 \cos 8 \theta}}}=$
- A$\sin 2 \theta$
- Correct answerB$2 \cos \theta$
- C$2 \sin \theta$
- D$2 \cos \frac{\theta}{2}$
Solution
Let $y=\sqrt{2+\sqrt{2+\sqrt{2+2 \cos 8 \theta}}}$
We know, $1+\cos 2 A=2 \cos ^2 A$
Therefore, $y=\sqrt{2+\sqrt{2+\sqrt{2 \cdot(1+\cos 8 \theta)}}}$
$$
\begin{aligned}
& =\sqrt{2+\sqrt{2+\sqrt{2 \cdot 2 \cos ^2 4 \theta}}} \\
& =\sqrt{2+\sqrt{2+2 \cos 4 \theta}}=\sqrt{2+\sqrt{2 \cdot 2 \cos ^2 2 \theta}} \\
& =\sqrt{2+2 \cos 2 \theta}=\sqrt{2 \cdot 2 \cos ^2 \theta}=2 \cos \theta
\end{aligned}
$$
Hence, the correct answer is option: 'B'.