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[KOPMOPAH]

Да се намери границата:$$\lim_{n \to \infty}\Big(\cos \frac{\pi}{4}\cos \frac{\pi}{8}\cos \frac{\pi}{16}...\cos \frac{\pi}{2^{n+1}}\Big)$$

[Добромир Глухаров]

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$A_n=cos\frac{\pi}{4}cos\frac{\pi}{8}cos\frac{\pi}{16}\cdots cos\frac{\pi}{2^{n+1}}$

$\boxed{2sin\frac{\pi}{2^{n+1}}cos\frac{\pi}{2^{n+1}}=sin\frac{\pi}{2^n}}$

$2sin\frac{\pi}{2^{n+1}}\cdot A_n=sin\frac{\pi}{2^n}\cdot A_{n-1}$

$\Rightarrow A_n=\frac{sin\frac{\pi}{2^n}}{2sin\frac{\pi}{2^{n+1}}}\cdot A_{n-1}=\frac{sin\frac{\pi}{2^n}}{2sin\frac{\pi}{2^{n+1}}}\cdot\frac{sin\frac{\pi}{2^{n-1}}}{2sin\frac{\pi}{2^n}}\cdot A_{n-2}$

$\cdots$

$\Rightarrow A_n=\frac{sin\frac{\pi}{2^2}}{2^{n-1}.sin\frac{\pi}{2^{n+1}}}\cdot A_1=\frac{sin\frac{\pi}{4}}{2^{n-1}.sin\frac{\pi}{2^{n+1}}}\cdot cos\frac{\pi}{4}$

$A_n=\frac{sin\frac{\pi}{2}}{2^n.sin\frac{\pi}{2^{n+1}}}=\frac{1.2}{2^{n+1}.sin\frac{\pi}{2^{n+1}}}=\frac{2}{\pi}\cdot\frac{\frac{\pi}{2^{n+1}}}{sin\frac{\pi}{2^{n+1}}}$

$\lim_{n\to\infty}A_n=\frac{2}{\pi}\cdot\frac{1}{\lim_{n\to\infty}\frac{sin\frac{\pi}{2^{n+1}}}{\frac{\pi}{2^{n+1}}}}=\frac{2}{\pi}$