How to find the limit of this product?

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Homework Help Overview

The discussion revolves around finding the limit of a product involving cosine functions, specifically the expression An = cos(pi/4)*cos(pi/8)*...*cos(pi/(2^n)). The original poster expresses confusion about how to approach the problem, particularly in relation to using sine functions in the context of a product of cosines.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the relationship between sine and cosine, particularly through the identity sin(2x) = 2*sin(x)*cos(x). There is an attempt to manipulate the expression for An using this identity, and questions arise about the meaning of "approaching infinity" in the context of limits.

Discussion Status

Some participants have provided hints and guidance on how to relate the product of cosines to sine functions, while others are clarifying terminology and concepts. The discussion is ongoing, with no explicit consensus reached yet.

Contextual Notes

The original poster indicates a lack of familiarity with products in contrast to sums, which may influence their understanding of the problem. There is also a language barrier noted by one participant, which could affect communication in the discussion.

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How to find the limit of this product?...

http://img55.imageshack.us/img55/8418/hellppppppppppcf7.png

http://img98.imageshack.us/img98/1153/help1fp4.png


the profesor gives us this problem but I don't know how to do it because he always work with sumatories but never with products, I want to know how can I use sine when the product is of cosine, help me please...
 
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OK, you have An = cos(pi/4)*cos(pi/8)*...*cos(pi/(2^n)).
The hint is that sin(2x) = 2*sin(x)*cos(x).

If you solve that equation for cos(x), you get cos(x) = sin(2x)/(2*sin(x))

Substitute this into each factor on the right side of the equation for An, so that An = [sin(pi/2)*sin(pi/4)*sin(pi/8)*...*sin(pi/(2^(n-1)))] / [2^(n-1)*sin(pi/4)*sin(pi/8)*...*sin(pi/(2^(n-1)))*sin(pi/(2^n))].

All but one factor in the numerator also appears in the denominator, so the expression for An can be simplified to what you show. All you have to do then is take the limit as n increases without bound.
 


what you mean with "without bound" (sorry my english isn't good)
 


It's another way to say that n is approaching infinity.
 


ah ok... thanks!... I will try to do it...
 

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