103 Trigonometry Problems

184 103 Trigonometry Problems
47. [China 2003, byYumin Huang] Let n be a fixed positive integer. Determine the
smallest positive real number λ such that for any θ 1 ,θ 2 ,...,θ n in the interval
(
0,
π
2
)
,if
tan θ 1 tan θ 2 ···tan θ n = 2 n/2 ,
then
cos θ 1 + cos θ 2 +···+cos θ n ≤ λ.
Solution: The answer is
⎧ √
3
⎪⎨ 3 , n = 1;
λ = 2 √ 3
3 ⎪⎩
, n = 2;
n − 1, n ≥ 3.
The case n = 1 is trivial. If n = 2, we claim that
cos θ 1 + cos θ 2 ≤ 2√ 3
3 ,
with equality if and only if θ 1 = θ 2 = tan −1 √ 2. It suffices to show that
cos 2 θ 1 + cos 2 θ 2 + 2 cos θ 1 cos θ 2 ≤ 4 3 ,
or
1
1 + tan 2 θ 1
+
√
1
1 + tan 2 + 2
θ 1
1
(
1 + tan 2 θ 1
)(
1 + tan 2 θ 2
) ≤ 4 3 .
Because tan θ 1 tan θ 2 = 2,
(1 + tan 2 θ 1
)(1 + tan 2 θ 2
)
= 5 + tan 2 θ 1 + tan 2 θ 2 .
By setting tan 2 θ 1 + tan θ 2 2
= x, the last inequality becomes
or
√
2 + x 1
5 + x + 2 5 + x ≤ 4 3 ,
√
1
2
5 + x ≤ 14 + x
3(5 + x) .
Squaring both sides and clearing denominators, we get 36(5 + x) ≤ 196 +
28x + x 2 , that is, 0 ≤ x 2 − 8x + 16 = (x − 4) 2 . This establishes our claim.