103 Trigonometry Problems

5. Solutions to Advanced Problems 185
Now assume that n ≥ 3. We claim that λ = n − 1. Note that λ ≥ n − 1; by
setting θ 2 = θ 3 =···=θ n = θ and letting θ → 0, we find that θ 1 → π 2 , and
so the left-hand side of the desired inequality approaches n − 1. It suffices to
show that
cos θ 1 + cos θ 2 +···+cos θ n ≤ n − 1.
Without loss of generality, assume that θ 1 ≥ θ 2 ≥···≥θ n . Then
It suffices to show that
tan θ 1 tan θ 2 tan θ 3 ≥ 2 √ 2.
cos θ 1 + cos θ 2 + cos θ 3 < 2.
(∗)
Because √ 1 − x 2 ≤ 1 − 1 2 x2 , cos θ i = √ 1 − sin 2 θ i < 1 − 1 2 sin2 θ i . Consequently,
by the arithmetic–geometric means inequality,
cos θ 2 + cos θ 3 < 2 − 1 )
(sin 2 θ 2 + sin 2 θ 3 ≤ 2 − sin θ 2 sin θ 3 .
2
Because
we have
or
tan 2 θ 1 ≥
8
tan 2 θ 2 tan 2 θ 3
,
sec 2 θ 1 ≥ 8 + tan2 θ 2 tan 2 θ 3
tan 2 θ 2 tan 2 θ 3
,
cos θ 1 ≤
tan θ 2 tan θ 3
sin θ 2 sin θ 3
√ = √ .
8 + tan 2 θ 2 tan 2 θ 3 8 cos 2 θ 2 cos 2 θ 3 + sin 2 θ 2 sin 2 θ 3
It follows that
cos θ 1 + cos θ 2 + cos θ 3
< 2 − sin θ 2 sin θ 3
[1 −
It is clear that the equality
]
1
8 cos 2 θ 2 cos 2 θ 3 + sin 2 θ 2 sin 2 .
θ 3
8 cos 2 θ 2 cos 2 θ 3 + sin 2 θ 2 sin 2 θ 3 ≥ 1, (∗∗)
establishes the desired inequality (∗). Inequality (∗∗) is equivalent to
8 + tan 2 θ 2 tan 2 θ 3 ≥ (1 + tan 2 θ 2 )(1 + tan 2 θ 3 ),