# A problem about sin and cos

1. Feb 19, 2005

### 3.14lwy

Given that

n = Π[2 – 2cos(kπ/ n)] ........ (where Π is the product sign , from k = 1 to n-1 )

as
cos2@ = 1 – 2(sin@)^2

then
2 – 2cos(kπ/ n) = 4[sin(kπ/ 2n)]^2 , for k = 1 , 2 , 3 , … n-1

then
n = Π[4[sin(kπ/ 2n)]^2] = [4^(n-1)] Π[sin(kπ/ 2n)]^2

but the book then said
Π[sin(kπ/ n)]^2 = n / [4^(n-1)]

why ?
why is not the sin(kπ/ 2n) but sin(kπ/ n) ?

2. Feb 19, 2005

### dextercioby

It's definitely a typo.It must be the "2" in the denominator.

Daniel.

3. Feb 19, 2005

### 3.14lwy

thank you first .

actually , the question is asking me to show :

$$\prod_{1}^{n-1} \sin{(\frac{k\pi}{n})} = \frac{n^{0.5}}{2^{n-1}}$$

by finding the roots of $$\ x^{2n} - 1 = 0$$

is the question wrong or I have made misstake?

4. Feb 20, 2005

### dextercioby

Again it's the "2" in the denominator missing...As for the equation,solve it and see whether you can relate the solutions to the identity which you have proven.

Daniel.