Geometry and Trigonometry 667n√
a+√
a^2 − 1 +
n√
a−√
a^2 − 1 =n√
cosht+sinht+n√
cosht−sinht=n√
et+n√
e−t=et/n+e−t/n=2 cosh
t
n.
It follows that coshntis rational. From the recurrence relation
cosh(k+ 1 )α=2 coshαcoshkα−cosh(k− 1 )α, k≥ 1 ,applied toα=tn, we can prove inductively that coshktnis rational for all positive integers
k. In particular, coshnnt=cosht=ais rational. This contradicts the hypothesis. Hence
our assumption was false and the conclusion follows.
(Romanian Team Selection Test for the International Mathematical Olympiad, 1979,
proposed by T. Andreescu)
688.We use the triple-angle formula
sin 3x=3 sinx−4 sin^3 x,which we rewrite as
sin^3 x=1
4
(3 sinx−sin 3x).The expression on the left-hand side of the identity from the statement becomes
27 ·
3 sin 9◦−sin 27◦
4+ 9 ·
3 sin 27◦−sin 81◦
4+ 3 ·
3 sin 81◦−sin 243◦
4
+3 sin 243◦−sin 729◦
4.
This collapses to
81 sin 9◦−sin 729◦
4=
81 sin 9◦−sin 9◦
4
=20 sin 9◦.(T. Andreescu)689.The triple-angle formula for the tangent gives
3 tan 3x=3 (3 tanx−tan^3 x)
1 −3 tan^2 x=
3 tan^3 x−9 tanx
3 tan^2 x− 1=tanx−8 tanx
3 tan^2 x− 1.
Hence
1
cotx−3 tanx=
tanx
1 −3 tan^2 x=
1
8
(3 tan 3x−tanx) for allx =k
π
2,k∈Z.