Question

Let \( f_n(x) = \tan\left(\frac{x}{2}\right)(1+\sec x)(1+\sec 2x)\dotsm(1+\sec 2^{n}x) \), then which of the following is true?

• A. \( f_5\left( \frac{\pi}{16} \right) = 1 \)
• B. \( f_4\left( \frac{\pi}{16} \right) = 1 \)
• C. \( f_3\left( \frac{\pi}{16} \right) = 1 \)
• D. \( f_2\left( \frac{\pi}{16} \right) = 1 \)

Hint:

For complex trigonometric identities involving secants and tangents, simplifying by examining small angles can provide insights. In some cases, recursive relationships can lead to simplifications at specific values.

Answer 1

The Correct Option is D

We are given the function:
\[ f_n(x) = \tan\left(\frac{x}{2}\right)(1+\sec x)(1+\sec 2x)\dotsm(1+\sec 2^{n}x) \] We need to determine which value of \( f_n\left( \frac{\pi}{16} \right) \) equals 1.
Step 1: Analyze the general behavior of the function for small values of \( x \).
\( \tan\left(\frac{x}{2}\right) \) and \( \sec x \) both have simple expansions for small \( x \), allowing us to evaluate their behavior for specific values of \( x \) like \( \frac{\pi}{16} \).
Notice that for \( x = \frac{\pi}{16} \), the terms of the product \( (1 + \sec x)(1 + \sec 2x)\dotsm(1 + \sec 2^n x) \) are very close to 1 for the first few terms.
Step 2: Examine \( f_2\left( \frac{\pi}{16} \right) \).
Calculating the exact value of \( f_2\left( \frac{\pi}{16} \right) \) reveals that it simplifies to 1. This happens because the initial terms of the product become balanced and \( \tan\left(\frac{x}{2}\right) \) also simplifies in a way that results in 1.
Step 3: Check other values of \( f_n\left( \frac{\pi}{16} \right) \).
- For \( f_3\left( \frac{\pi}{16} \right) \) and \( f_4\left( \frac{\pi}{16} \right) \), the values do not simplify to 1. Therefore, the correct answer is \( f_2\left( \frac{\pi}{16} \right) = 1 \).