Advanced book on Mathematics Olympiad

668 Geometry and Trigonometry

It follows that the left-hand side telescopes as

1

8

(3 tan 27◦−tan 9◦+9 tan 81◦−3 tan 27◦+27 tan 243◦−9 tan 81◦+81 tan 729◦

−27 tan 243◦)=

1

8

(81 tan 9◦−tan 9◦)=10 tan 9◦.

(T. Andreescu)

690.Multiply the left-hand side by sin 1◦and transform it using the identity

sin((k+ 1 )◦−k◦)

sink◦sin(k+ 1 )◦

=cotk◦−cot(k+ 1 )◦.

We obtain

cot 45◦−cot 46◦+cot 47◦−cot 48◦+···+cot 131◦−cot 132◦+cot 133◦−cot 134◦.

At first glance this sum does not seem to telescope. It does, however, after changing

the order of terms. Indeed, if we rewrite the sum as

cot 45◦−(cot 46◦+cot 134◦)+(cot 47◦+cot 133◦)−(cot 48◦+cot 132◦)

+···+(cot 89◦+cot 91◦)−cot 90◦,

then the terms in the parentheses cancel, since they come from supplementary angles.

The conclusion follows from cot 45◦=1 and cot 90◦=0.

(T. Andreescu)

691.The formula

tan(a−b)=

tana−tanb

1 +tanatanb

translates into

arctan

x−y

1 +xy

=arctanx−arctany.

Applied tox=n+1 andy=n−1, it gives

arctan

2

n^2

=arctan

(n+ 1 )−(n− 1 )

1 +(n+ 1 )(n− 1 )

=arctan(n+ 1 )−arctan(n− 1 ).

The sum in part (a) telescopes as follows:

∑∞

n= 1

arctan

2

n^2

= lim

N→∞

∑N

n= 1

arctan

2

n^2

= lim

N→∞

∑N

n= 1

(arctan(n+ 1 )−arctan(n− 1 ))