103 Trigonometry Problems

132 103 Trigonometry Problems
and so the range of c is the interval [− √ 2, √ 2]. Consequently, it suffices to
find the minimum of
P(c) =
2ab(a + b) + 2 + 2(a + b)
∣ 2ab ∣
=
c(c 2 ∣
− 1) + 2(c + 1)
∣∣∣ ∣ c 2 − 1 ∣ = c + 2
c − 1∣
=
∣ c − 1 + 2 ∣ ∣∣∣
c − 1 + 1 .
for c in the interval [− √ 2, √ 2].Ifc−1 > 0, then by the arithmetic–geometric
means inequality, (c−1)+
c−1 2 > 2√ 2, and so P(c) > 1+2 √ 2. If c−1 < 0,
then by the same token,
(c − 1) + 2 (
c − 1 =− (1 − c) + 2 )
≤−2 √ 2,
1 − c
with equality if and only if 1 − c =
1−c 2 ,orc = 1 − √ 2. It follows that the
∣
minimum value sought is ∣−2 √ 2 + 1∣ = 2 √ 2−1, obtained when c = 1− √ 2.
Note: Taking the derivative of the function
f(x)= sin x + cos x + tan x + cot x + sec x + csc x
and considering only its critical points is a troublesome approach to this problem,
because it is difficult to show that f(x)does not cross the x axis smoothly.
Indeed, with a little bit more work, we can show that f(x) ̸= 0 with the presented
solution.
10. [Belarus 1999] Two real sequences x 1 ,x 2 ,... and y 1 ,y 2 ,... are defined in
the following way:
x 1 = y 1 = √ √
3, x n+1 = x n + 1 + xn 2, y y n
n+1 =
1 + √ ,
1 + yn
2
for all n ≥ 1. Prove that 2 1.
Solution: Writing x n = tan a n for 0 ◦