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

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SUMMARY

The function \( f_n(\theta) \) for positive integers \( n \) is defined as \( f_n(\theta) = \tan \frac{\theta}{2}(1+\sec \theta)(1+\sec 2\theta)(1+\sec 4 \theta)\cdots (1+\sec2^n \theta) \). Through mathematical induction, it is established that \( f_n(\theta) = \tan(2^n \theta) \). Consequently, for \( n = 2, 3, 4, 5 \) and respective angles \( \frac{\pi}{16}, \frac{\pi}{32}, \frac{\pi}{64}, \frac{\pi}{128} \), the values of \( f_n \) are all equal to \( \tan\left(\frac{\pi}{4}\right) = 1 \).

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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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