How to find the limit of this product?

[medinap]

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...

 

[Mark44]

OK, you have A_n = 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 A_n, so that A_n = [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 A_n can be simplified to what you show. All you have to do then is take the limit as n increases without bound.

 

[medinap]

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

 

[Mark44]

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

 

[medinap]

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