KCET · Maths · Trigonometric Ratios & Identities
\(\sqrt{2+\sqrt{2+\sqrt{2+2 \cos 8 \theta}}}=\)
- A \(\sin 2 \theta\)
- B \(2 \cos \theta\)
- C \(2 \sin \theta\)
- D \(2 \cos \frac{\theta}{2}\)
Answer & Solution
Correct Answer
(B) \(2 \cos \theta\)
Step-by-step Solution
Detailed explanation
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, \(y=\sqrt{2+\sqrt{2+\sqrt{2 \cdot(1+\cos 8 \theta)}}}\)
\[
\begin{aligned}
& =\sqrt{2+\sqrt{2+\sqrt{2 \cdot 2 \cos ^2 4 \theta}}} \\
& =\sqrt{2+\sqrt{2+2 \cos 4 \theta}}=\sqrt{2+\sqrt{2 \cdot 2 \cos ^2 2 \theta}} \\
& =\sqrt{2+2 \cos 2 \theta}=\sqrt{2 \cdot 2 \cos ^2 \theta}=2 \cos \theta
\end{aligned}
\]
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