Advanced High-School Mathematics

SECTION 5.4 Polynomial Approximations 297

(i) By equating the imaginary parts of DeMoivre’s formula

cosnθ+isinnθ= (cosθ+isinθ)n= sinnθ(cotθ+i)n,

obtain the identity

sinnθ= sinnθ







Ñ

n

1

é

cotn−^1 θ−

Ñ

n

3

é

cotn−^3 θ+

Ñ

n

5

é

cotn−^5 θ−···





.

(ii) Letn= 2m+ 1 and express the above as

sin(2m+ 1)θ= sin^2 m+1θPm(cot^2 θ), 0 < θ <

π

2

,

wherePm(x) is the polynomial of degreemgiven by

Pm(x) =

Ñ

2 m+ 1

1

é

xm−

Ñ

2 m+ 1

3

é

xm−^1 +

Ñ

2 m+ 1

5

é

xm−^2 −···.

(iii) Conclude that the real numbers

xk= cot^2

(

kπ

2 m+ 1

)

, 1 ≤k≤m,

are zeros of Pm(x), and that they are all distinct.

Therefore,x 1 , x 2 , ..., xmcompriseallof the zeros ofPm(x).

(iv) Conclude from part (iii) that

∑m

k=1

cot^2

( kπ

2 m+ 1

)

=

∑m

k=1

xk=

Ñ

2 m+ 1

3

é¬Ñ

2 m+ 1

1

é

=

m(2m−1)

3

,

proving the claim of Step 1.

Step 2. Starting with the familiar inequality sinx < x < tanx

for 0< x < π/2, show that

cot^2 x <

1

x^2

<1 + cot^2 x, 0 < x <

π

2

.

Step 3. Putx =

kπ

2 m+ 1

, where k andm are positive integers

and 1≤k≤m, and infer that

∑m

k=1

cot^2

( kπ

2 m+ 1

)

<

(2m+ 1)^2

π^2

∑m

k=1

1

k^2

< m+

∑m

k=1

cot^2

( kπ

2 m+ 1

)

.