103 Trigonometry Problems

4. Solutions to Introductory Problems 103
from which our desired result follows.
28. Let ABC be a triangle. Prove that
(a) cos A cos B cos C ≤ 1 8 ;
(b) sin A sin B sin C ≤ 3√ 3
8 ;
(c) sin A + sin B + sin C ≤ 3√ 3
2 .
(d) cos 2 A + cos 2 B + cos 2 C ≥ 3 4 ;
(e) sin 2 A + sin 2 B + sin 2 C ≤ 9 4 ;
(f) cos 2A + cos 2B + cos 2C ≥− 3 2 ;
(g) sin 2A + sin 2B + sin 2C ≤ 3√ 3
2 .
Solution: For part (a), if triangle ABC is nonacute, the left-hand side of the
inequality is nonpositive, and so the inequality is clearly true.
If ABC is acute, then cos A, cos B, cos C are all positive. To establish (a) and
(d), we need only note that the relation between (a) and (d) and Problem 24(d)
is similar to that of Problem 23(a) and (b) and Problem 22. (Please see the
note after the solution of Problem 23.)
The two inequalities in parts (d) and (e) are equivalent because cos 2 x +
sin 2 x = 1.
By (e) and by the arithmetic–geometric means inequality, wehave
9
4 ≥ sin2 A + sin 2 B + sin 2 C ≥ 3 3√ sin 2 A sin 2 B sin 2 C,
from which (b) follows.
From (a−b) 2 +(b−c) 2 +(c−a) 2 ≥ 0 or by application of Cauchy–Schwarz
inequality, we can show that 3 ( a 2 + b 2 + c 2) ≥ (a + b + c) 2 . By (e) and by
setting a = sin A, b = sin B, c = sin C, we obtain (c).
Part (f) follows from (e) and cos 2x = 2 cos 2 x − 1. Finally, (g) follows from
(b) and the identity
sin 2A + sin 2B + sin 2C = 4 sin A sin B sin C