The Real And Complex Number Systems

Proof: By Exercise 1.44 (i), we have
sin nθ =
[ n+1
2 ]
∑
k=1
( n2k−1
)
sin 2k−1 θ cos n−(2k−1) θ
⎧
⎫
⎪⎨ [ n+1
2
∑
]
(
= sin n θ
n2k−1
)
⎪⎬
cot n−(2k−1) θ
⎪⎩
⎪⎭
k=1
= sin n θ { ( n 1) cot n−1 θ − ( n 3) cot n−3 θ + ( n 5) cot n−5 θ − +... } .
(b) If 0 < θ < π/2, prove that
sin (2m + 1) θ = sin 2m+1 θP m
(
cot 2 θ )
where P m is the polynomial of degree m given by
P m (x) = ( )
2m+1
1 x m − ( )
2m+1
3 x m−1 + ( )
2m+1
5 x m−2 − +...
Use this to show that P m has zeros at the m distinct points x k = cot 2 {πk/ (2m + 1)}
for k = 1, 2, ..., m.
Proof: By (a),
sin (2m + 1) θ
= sin 2m+1 θ
{ (2m+1
1
) (
cot 2 θ ) m
−
( 2m+1
3
= sin 2m+1 θP m
(
cot 2 θ ) , where P m (x) =
) (
cot 2 θ ) m−1 (
+
2m+1
) (
5 cot 2 θ ) }
m−2
− +...
m+1
∑
k=1
( 2m+1
)
2k−1 x m+1−k . (*)
In addition, by (*), sin (2m + 1) θ = 0 if, and only if, P m (cot 2 θ) = 0. Hence,
P m has zeros at the m distinct points x k = cot 2 {πk/ (2m + 1)} for k =
1, 2, ..., m.
(c) Show that the sum of the zeros of P m is given by
m∑
cot 2 πk
2m + 1
k=1
34
m (2m − 1)
= ,
3