Question

Let \( P = \{ \theta \in [0, 4\pi] : \tan^2\theta \neq 1 \} \) and \( S = \{ a \in \mathbb{Z} : 2(\cos^8\theta - \sin^8\theta) \sec 2\theta = a^2, \theta \in P \} \). Then \( n(S) \) is:

• A. 0
• B. 1
• C. 2
• D. 3

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
We are dealing with a trigonometric identity that needs simplification.
After simplifying the expression \(2(\cos^8\theta - \sin^8\theta)\sec 2\theta\), we will find its range of values over the allowed domain of \(\theta\).
Since \(a\) must be an integer, \(a^2\) must be a perfect square integer that falls within this range.
Step 2: Key Formula or Approach:
We use the difference of squares factorization: \(x^4 - y^4 = (x^2 - y^2)(x^2 + y^2)\).
Apply this to \(\cos^8\theta - \sin^8\theta = (\cos^4\theta - \sin^4\theta)(\cos^4\theta + \sin^4\theta)\).
Further, \(\cos^4\theta - \sin^4\theta = (\cos^2\theta - \sin^2\theta)(\cos^2\theta + \sin^2\theta)\).
Recall that \(\cos^2\theta + \sin^2\theta = 1\) and \(\cos^2\theta - \sin^2\theta = \cos 2\theta\).
Step 3: Detailed Explanation:
Let's simplify the given expression for \(a^2\).
\[ a^2 = 2(\cos^8\theta - \sin^8\theta)\sec 2\theta \] \[ a^2 = 2(\cos^4\theta - \sin^4\theta)(\cos^4\theta + \sin^4\theta) \left(\frac{1}{\cos 2\theta}\right) \] \[ a^2 = 2(\cos^2\theta - \sin^2\theta)(\cos^2\theta + \sin^2\theta)(\cos^4\theta + \sin^4\theta) \left(\frac{1}{\cos 2\theta}\right) \] Substitute the basic identities.
\[ a^2 = 2(\cos 2\theta)(1)(\cos^4\theta + \sin^4\theta) \left(\frac{1}{\cos 2\theta}\right) \] Since \(\theta \in P \implies \tan^2\theta \neq 1 \implies \cos 2\theta \neq 0\), we can safely cancel \(\cos 2\theta\).
\[ a^2 = 2(\cos^4\theta + \sin^4\theta) \] We can rewrite \(\cos^4\theta + \sin^4\theta\) by completing the square.
\[ \cos^4\theta + \sin^4\theta = (\cos^2\theta + \sin^2\theta)^2 - 2\sin^2\theta \cos^2\theta = 1 - \frac{1}{2}(4\sin^2\theta \cos^2\theta) = 1 - \frac{1}{2}\sin^2 2\theta \] Substitute this back into the equation for \(a^2\).
\[ a^2 = 2\left(1 - \frac{1}{2}\sin^2 2\theta\right) = 2 - \sin^2 2\theta \] Now we determine the range of this expression.
We know that \(0 \le \sin^2 2\theta \le 1\).
However, the domain condition \(\tan^2\theta \neq 1\) means \(\theta \neq \frac{\pi}{4}, \frac{3\pi}{4}, \dots\), which implies \(2\theta \neq \frac{\pi}{2}, \frac{3\pi}{2}, \dots\).
Therefore, \(\sin^2 2\theta\) can never be exactly equal to \(1\).
So the range of \(\sin^2 2\theta\) is \([0, 1)\).
This means the range of \(a^2\) is \(2 - [0, 1) = (1, 2]\).
We need \(a \in \mathbb{Z}\), which means \(a^2\) must be a perfect square integer.
The only integer in the interval \((1, 2]\) is \(2\).
If \(a^2 = 2\), then \(a = \pm \sqrt{2}\), which are not integers.
Since there is no integer \(a\) that satisfies the equation, the set \(S\) is empty.
Step 4: Final Answer:
The number of elements \(n(S)\) is \(0\).