Question

Let \[ f_n(x) = \tan\frac{x}{2}(1+\sec x)(1+\sec 2x)\cdots(1+\sec 2^{n-1}x), \] then:

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

Hint:

For trig telescoping products: \begin{itemize} \item Convert sec terms into cosine ratios. \item Look for cancellation patterns. \end{itemize}

Answer 1

The Correct Option is B

Concept: Use identity: \[ 1 + \sec \theta = \frac{2\cos^2(\theta/2)}{\cos\theta} \] And telescoping products. Step 1: {\color{red}Rewrite terms.} Using: \[ 1+\sec\theta = \frac{2}{1+\cos\theta} \] Each term simplifies ratios of cosines. Step 2: {\color{red}Telescoping pattern.} Product becomes: \[ f_n(x) = \tan\frac{x}{2} \cdot \frac{\cos(x/2)}{\cos x} \cdot \frac{\cos x}{\cos 2x} \cdots \frac{\cos(2^{n-2}x)}{\cos(2^{n-1}x)} \] Most terms cancel: \[ f_n(x) = \frac{\sin(x/2)}{\cos(2^{n-1}x)} \] Step 3: {\color{red}Substitute \( x = \frac{\pi}{16} \).} \[ f_n = \frac{\sin(\pi/32)}{\cos\left(\frac{2^{n-1}\pi}{16}\right)} \] We want denominator = numerator. Set: \[ 2^{n-1}\frac{\pi}{16} = \frac{\pi}{2} \] \[ 2^{n-1} = 8 \Rightarrow n = 4 \] Step 4: {\color{red}Conclusion.} \[ f_4\!\left(\frac{\pi}{16}\right) = 1 \]