Advanced book on Mathematics Olympiad

658 Geometry and Trigonometry

668.Denote the sum in question byS 1 and let

S 2 =

(

n

1

)

sinx+

(

n

2

)

sin 2x+···+

(

n

n

)

sinnx.

Using Euler’s formula, we can write

1 +S 1 +iS 2 =

(

n

0

)

+

(

n

1

)

eix+

(

n

2

)

e^2 ix+···+

(

n

n

)

einx.

By the multiplicative property of the exponential we see that this is equal to

∑n

k= 0

(

n

k

)(

eix

)k

=

(

1 +eix

)n

=

(

2 cos

x

2

)n(

ei

x 2 )n

.

The sum in question is the real part of this expression less 1, which is equal to

2 ncosn

x

2

cos

nx

2

− 1.

669.Combinef(x)with the functiong(x)=excosθsin(xsinθ)and write

f(x)+ig(x)=excosθ(cos(xsinθ)+isin(xsinθ))

=excosθ·eixsinθ=ex(cosθ+isinθ).

Using the de Moivre formula we expand this in a Taylor series as

1 +

x

1!

(cosθ+isinθ)+

x^2

2!

(cos 2θ+isin 2θ)+···+

xn

n!

(cosnθ+isinnθ )+···.

Consequently, the Taylor expansion off(x)around 0 is the real part of this series, i.e.,

f(x)= 1 +

cosθ

1!

x+

cos 2θ

2!

x^2 +···+

cosnθ

n!

xn+···.

670.Letzj=r(costj+isintj), withr =0 andtj∈( 0 ,π)∪(π, 2 π),j= 1 , 2 ,3. By
hypothesis,

sint 1 +rsin(t 2 +t 3 )= 0 ,

sint 2 +rsin(t 3 +t 1 )= 0 ,

sint 3 +rsin(t 1 +t 2 )= 0.

Lett=t 1 +t 2 +t 3. Then

sintj=−rsin(t−ti)=−rsintcostj−rcostsintj, forj= 1 , 2 , 3 ,