Cristian Calude, Problema 1.

[Laurian Filip]

a) Sa se arate ca \( \prod_{k=1}^{n-1}cos\frac{a}{2^k}=\frac{\sin a}{2^{n-1}sin \frac{a}{2^{n-1}} \), \( \forall n\geq 2, n \in \mathbb{N} \) si \( \forall a \in (0; \pi). \)

b) Sa se arate ca \( \frac{1}{{2^{n-1} \cdot sin{\frac{a}{2^{n-1}}}} }=\frac{\prod_{i=n-k+1}^{n-1} cos{\frac{a}{2^i}}} {2^{n-k} \cdot sin {\frac{a}{2^{n-k}}} \)

c) Se considera sirul \( (x)_{n\geq1} \) definit astfel: \( x_1=1 \), \( x_{n+1}=cos {\frac{a}{2^n}} \cdot x_n + \frac{sina}{2^n \cdot sin {\frac{a}{2^n}} \), \( \forall n\geq1 \), unde \( a \in (0; \pi) \).
Sa se calculeze \( \lim_{n \rightarrow \infty} \frac{x_n}{n} \)