# How to find the limit of this product?

1. Oct 25, 2008

### medinap

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

the profesor gives us this problem but I dont 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...

2. Oct 25, 2008

### Staff: Mentor

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

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.

3. Oct 25, 2008

### medinap

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

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

4. Oct 25, 2008

### Staff: Mentor

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

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

5. Oct 25, 2008

### medinap

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

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