अऋण पूर्णांकों n के लिए,f(n) = frac{sum_{k=0} | vedclass

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(B) सर्वसमिका $2 \sin A \sin B = \cos(A-B) - \cos(A+B)$ और $2 \sin^2 A = 1 - \cos(2A)$ का उपयोग करते हुए:$f(n) = \frac{\sum_{k=0}^n [\cos(\frac{\pi}{n

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$1, 2, 4$

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(B) सर्वसमिका $2 \sin A \sin B = \cos(A-B) - \cos(A+B)$ और $2 \sin^2 A = 1 - \cos(2A)$ का उपयोग करते हुए:$f(n) = \frac{\sum_{k=0}^n [\cos(\frac{\pi}{n+2}) - \cos(\frac{2k+3}{n+2}\pi)]}{\sum_{k=0}^n [1 - \cos(\frac{2k+2}{n+2}\pi)]}$चूंकि $\sum_{k=0}^n \cos(\frac{2k+2}{n+2}\pi) = 0$ और $\sum_{k=0}^n \cos(\frac{2k+3}{n+2}\pi) = -\cos(\frac{\pi}{n+2})$,व्यंजक का सरलीकरण:$f(n) = \frac{(n+1) \cos(\frac{\pi}{n+2}) + \cos(\frac{\pi}{n+2})}{n+1} = \cos(\frac{\pi}{n+2})$.$(1)$ $f(5) = \cos(\frac{\pi}{7}) \implies \sin(7 \cos^{-1} f(5)) = \sin(7 \cdot \frac{\pi}{7}) = \sin(\pi) = 0$. (सही)$(2)$ $f(4) = \cos(\frac{\pi}{6}) = \frac{\sqrt{3}}{2}$. (सही)$(3)$ $\lim_{n ightarrow \infty} f(n) = \lim_{n ightarrow \infty} \cos(\frac{\pi}{n+2}) = \cos(0) = 1 
eq \frac{1}{2}$. (गलत)$(4)$ $f(6) = \cos(\frac{\pi}{8}) \implies \alpha = 	an(\frac{\pi}{8}) = \sqrt{2}-1$. तब $\alpha^2 + 2\alpha - 1 = (\sqrt{2}-1)^2 + 2(\sqrt{2}-1) - 1 = (2 - 2\sqrt{2} + 1) + (2\sqrt{2} - 2) - 1 = 0$. (सही)अतः,विकल्प $1, 2, 4$ सही हैं।

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