are given by x^2 − 2 xcos(θ+^2 rπ
n)+1 for r= 0, 1 , 2 ,.. ., n − 1. This implies x^2 n− 2 xncos nθ +1
can be expressed as
x^2 n− 2 xncos nθ + 1 =n∏− 1r=0(x^2 − 2 xcos(θ+^2 rπ
n) + 1) (12.130)In the special case x= 1 the equation (12.130) simplifies to
1 −cos nθ = 2n−^1n∏− 1r=0(
1 −cos(
θ+^2 rπ
n))
(12.131)Replacing θby 2 θin equation (12.131) and simplifying one obtains
2 sin^2 nθ = 2n−^12 nsin^2 θsin^2 (θ+π
n) sin^2 (θ+^2 π
n)···sin^2 (θ+(n−1)π
n) (12.132)Further simplify equation (12.132) and then take the square root of both sides to
obtain
sin nθ
sin θ = 2n− (^1) sin(θ+π
n) sin(θ+
2 π
n)···sin(θ+
(n−1)π
n ) (12.133)
where the positive square root is taken when each term is positive. In equation
(12.133) take the limit as θ→ 0 and verify
n= 2n−^1 sin( π
n) sin(^2 π
n)···sin( (n−1)π
n) (12.134)Now consider the product
y= Γ(
1
n)
Γ(
2
n)
Γ(
3
n)
Γ(
4
n)
···Γ(
n− 1
n)and reverse the terms within the product to show
y^2 =[
Γ(
1
n)
Γ(
n− 1
n)][
Γ(
2
n)
Γ(
n− 2
n)]
···[
Γ(
n− 1
n)
Γ(
1
n)]followed by writing equation (12.112) in the form
Γ(m
n)
Γ(
n−m
n)
=π
sin( mπn)for m= 1, 2 ,.. ., n − 1. This identity produces
y^2 =π
sin( nπ)π
sin(^2 nπ)π
sin(^3 nπ)···π
sin((n−n1)π)One can now employ the result from Example 12-6, equation (12.134), to show
Γ(
1
n)
Γ(
2
n)
Γ(
3
n)
Γ(
4
n)
···Γ(
n− 1
n)
=(2π)n−^21
n^12