MHB Calculating $f_n(\theta)$ for Positive Integers $n$

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The function \( f_n(\theta) \) is defined as \( f_n(\theta) = \tan \frac{\theta}{2}(1+\sec \theta)(1+\sec 2\theta)(1+\sec 4 \theta)\cdots (1+\sec2^n \theta) \). It has been proven through mathematical induction that \( f_n(\theta) = \tan(2^n \theta) \) for positive integers \( n \). Consequently, the values for \( f_2 \left(\frac{\pi}{16}\right) \), \( f_3 \left(\frac{\pi}{32}\right) \), \( f_4 \left(\frac{\pi}{64}\right) \), and \( f_5 \left(\frac{\pi}{128}\right) \) all equal \( \tan\left(\frac{\pi}{4}\right) \), which is 1. This result simplifies the calculations for these specific cases. The discussion highlights the elegant relationship between the function and the tangent of doubled angles.
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For a positive integer $n$, let

$$f_n(\theta)=\tan \frac{\theta}{2}(1+\sec \theta)(1+\sec 2\theta)(1+\sec 4 \theta)\cdots (1+\sec2^n \theta)$$

Find the value of

(i) $f_2 \left(\dfrac{\pi}{16} \right)$

(ii) $f_3 \left(\dfrac{\pi}{32} \right)$

(iii) $f_4 \left(\dfrac{\pi}{64} \right)$

(iv) $f_5 \left(\dfrac{\pi}{128} \right)$
 
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sbhatnagar said:
For a positive integer $n$, let

$$f_n(\theta)=\tan \frac{\theta}{2}(1+\sec \theta)(1+\sec 2\theta)(1+\sec 4 \theta)\cdots (1+\sec2^n \theta)$$

Find the value of

(i) $f_2 \left(\dfrac{\pi}{16} \right)$

(ii) $f_3 \left(\dfrac{\pi}{32} \right)$

(iii) $f_4 \left(\dfrac{\pi}{64} \right)$

(iv) $f_5 \left(\dfrac{\pi}{128} \right)$

Hi sbhatnagar, :)

It can be shown by mathematical induction that,

\[f_n(\theta)=\tan{2^{n}\theta}\mbox{ where }n\in\mathbb{Z}^{+}\]

Therefore,

\[f_2 \left(\dfrac{\pi}{16} \right)=f_3 \left(\dfrac{\pi}{32} \right)=f_4 \left(\dfrac{\pi}{64} \right)=f_5 \left(\dfrac{\pi}{128} \right)=\tan\left(\frac{\pi}{4}\right)=1\]

Kind Regards,
Sudharaka.
 
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