103 Trigonometry Problems

5. Solutions to Advanced Problems 145
( )
It suffices to show that b n = cot 2 n−3 π
3
, where b n = a n + 2, n ≥ 1. The
recursive relation becomes
or
b n+1 − 2 = (b n − 2) 2 − 5
2b n
,
b n+1 = b2 n − 1
2b n
.
Assuming, inductively, that b k = cot c k , where c k = 2k−3 π
3
, yields
and we are done.
b k+1 = cot2 c k − 1
2 cot c k
= cot 2c k = cot c k+1 ,
23. [APMC 1982] Let n be an integer with n ≥ 2. Prove that
n∏
k=1
[ )]
π
tan
(1 + 3k
3 3 n =
− 1
n∏
k=1
[ )]
π
cot
(1 − 3k
3 3 n .
− 1
Solution: Let
[ )]
π
u k = tan
(1 + 3k
3 3 n − 1
The desired equality becomes
Set
[ )]
π
and v k = tan
(1 − 3k
3 3 n .
− 1
n∏
u k v k = 1.
k=1
t k = tan 3k−1 π
3 n − 1 .
Applying the addition and and subtraction formulas yields
u k = tan
( )
π
3 + 3k−1 π
3 n − 1
The triple-angle formulas give
=
√ √
3 + tk
3 −
1 − √ tk
and v k =
3t k 1 + √ .
3t k
(∗)
t k+1 = 3t k − tk
3
1 − 3tk
2 ,