Q118 / 180 · COMEDK 2024 (Evening Shift) · Mathematics
$\sqrt{2+\sqrt{2+\sqrt{2+2 \cos 8 \theta}}} \text { where } \theta \in\left[-\dfrac{\pi}{8}, \dfrac{\pi}{8}\right] \text { is equal to }$
- A$\cos 2 \theta$
- B$\sin 2 \theta$
- C$2 \sin \theta$
- Correct answerD$2 \cos \theta$
Solution
Let the given expression be $E = \sqrt{2+\sqrt{2+\sqrt{2+2 \cos 8 \theta}}}$.
Using the trigonometric identity $1 + \cos 2A = 2 \cos^2 A$, we have $2 + 2 \cos 8 \theta = 2(1 + \cos 8 \theta) = 2(2 \cos^2 4 \theta) = 4 \cos^2 4 \theta$.
Since $\theta \in [-\dfrac{\pi}{8}, \dfrac{\pi}{8}]$, $4 \theta \in [-\dfrac{\pi}{2}, \dfrac{\pi}{2}]$, so $\cos 4 \theta \ge 0$. Thus, $\sqrt{2+2 \cos 8 \theta} = \sqrt{4 \cos^2 4 \theta} = 2 \cos 4 \theta$.
Substituting this into the expression, $E = \sqrt{2+\sqrt{2+2 \cos 4 \theta}}$.
Applying the same identity again, $2 + 2 \cos 4 \theta = 2(1 + \cos 4 \theta) = 2(2 \cos^2 2 \theta) = 4 \cos^2 2 \theta$.
Since $2 \theta \in [-\dfrac{\pi}{4}, \dfrac{\pi}{4}]$, $\cos 2 \theta \ge 0$. Thus, $\sqrt{2+2 \cos 4 \theta} = 2 \cos 2 \theta$.
Substituting this into the expression, $E = \sqrt{2+2 \cos 2 \theta}$.
Applying the identity once more, $2 + 2 \cos 2 \theta = 2(1 + \cos 2 \theta) = 2(2 \cos^2 \theta) = 4 \cos^2 \theta$.
Since $\theta \in [-\dfrac{\pi}{8}, \dfrac{\pi}{8}]$, $\cos \theta \ge 0$. Thus, $E = \sqrt{4 \cos^2 \theta} = 2 \cos \theta$.
Answer: $2 \cos \theta$
Using the trigonometric identity $1 + \cos 2A = 2 \cos^2 A$, we have $2 + 2 \cos 8 \theta = 2(1 + \cos 8 \theta) = 2(2 \cos^2 4 \theta) = 4 \cos^2 4 \theta$.
Since $\theta \in [-\dfrac{\pi}{8}, \dfrac{\pi}{8}]$, $4 \theta \in [-\dfrac{\pi}{2}, \dfrac{\pi}{2}]$, so $\cos 4 \theta \ge 0$. Thus, $\sqrt{2+2 \cos 8 \theta} = \sqrt{4 \cos^2 4 \theta} = 2 \cos 4 \theta$.
Substituting this into the expression, $E = \sqrt{2+\sqrt{2+2 \cos 4 \theta}}$.
Applying the same identity again, $2 + 2 \cos 4 \theta = 2(1 + \cos 4 \theta) = 2(2 \cos^2 2 \theta) = 4 \cos^2 2 \theta$.
Since $2 \theta \in [-\dfrac{\pi}{4}, \dfrac{\pi}{4}]$, $\cos 2 \theta \ge 0$. Thus, $\sqrt{2+2 \cos 4 \theta} = 2 \cos 2 \theta$.
Substituting this into the expression, $E = \sqrt{2+2 \cos 2 \theta}$.
Applying the identity once more, $2 + 2 \cos 2 \theta = 2(1 + \cos 2 \theta) = 2(2 \cos^2 \theta) = 4 \cos^2 \theta$.
Since $\theta \in [-\dfrac{\pi}{8}, \dfrac{\pi}{8}]$, $\cos \theta \ge 0$. Thus, $E = \sqrt{4 \cos^2 \theta} = 2 \cos \theta$.
Answer: $2 \cos \theta$
Hence, the correct answer is option: 'D'.