103 Trigonometry Problems

28 103 Trigonometry Problems
Example 1.11. Let x,y,zbe positive real numbers such that x+y+z = 1. Determine
the minimum value of
1
x + 4 y + 9 z .
Solution: An application of Cauchy–Schwarz inequality makes this is a one-step
problem. Nevertheless, we present a proof which involves only the easier inequality
x 2 + y 2 ≥ 2xy for real numbers x and y, by setting first x = tan b and y = 2 tan b
and second x = tan a and y = cot a.
Clearly, z is a real number in the interval [0, 1]. Hence there is an angle a such that
z = sin 2 a. Then x + y = 1 − sin 2 a = cos 2 x
a,or
cos 2 a + y = 1. For an angle b,
cos 2 a
we have cos 2 b + sin 2 b = 1. Hence, we can set x = cos 2 a cos 2 b, y = cos 2 a sin 2 b
for some angle b. It suffices to find the minimum value of
or
P = sec 2 a sec 2 b + 4 sec 2 a csc 2 b + 9 csc 2 a,
P = (tan 2 a + 1)(tan 2 b + 1) + 4(tan 2 a + 1)(cot 2 b + 1) + 9(cot 2 a + 1).
Expanding the right-hand side gives
P = 14 + 5 tan 2 a + 9 cot 2 a + (tan 2 b + 4 cot 2 b)(1 + tan 2 a)
( )
≥ 14 + 5 tan 2 a + 9 cot 2 a + 2 tan b · 2 cot b 1 + tan 2 a
= 18 + 9(tan 2 a + cot 2 a) ≥ 18 + 9 · 2 tan a cot a = 36.
Equality holds when tan a = cot a and tan b = 2 cot b, which implies that cos 2 a =
sin 2 a and 2 cos 2 b = sin 2 b. Because sin 2 θ + cos 2 θ = 1, equality holds when
cos 2 a = 1 2 and cos2 b = 1 3 ; that is, x = 1 6 , y = 1 3 , z = 1 2 .
Ceva’s Theorem
A cevian of a triangle is any segment joining a vertex to a point on the opposite side.
[Ceva’s Theorem] Let AD, BE, CF be three cevians of triangle ABC. The following
are equivalent (see Figure 1.28):
(1) AD, BE, CF are concurrent; that is, these lines pass a common point;
(2) sin ̸ ABE
sin ̸ DAB · sin ̸ BCF
sin ̸ EBC · sin ̸ CAD
sin ̸ FCA = 1;