103 Trigonometry Problems

4. Solutions to Introductory Problems 97
Combining the last two equalities gives part (a).
Part (b) then follows from (a) and Problem 22. Part (c) then follows from part
(b) by noting that 1 − sin 2 x = cos 2 x. Finally, by (c) and by the arithmetic–
geometric means inequality, we have
9
4 ≥ A cos2 2 + B cos2 2 + C √
cos2 2 ≥ 3 3 cos 2 A B C 2 cos2 2 cos2 2 ,
implying (d).
Again by Problem 8, we have
csc A 2 ≥ b + c
a
= b a + c a
and analogous formulas for csc B 2 and csc C 2
. Then part (e) follows routinely
from the arithmetic–geometric means inequality.
Note: We present another approach to part (a). Note that sin A 2 , sin B 2 , sin C 2
are all positive. Let t = 3 √sin A 2 sin B 2 sin C 2 . It suffices to show that t ≤ 1 2 .By
the arithmetic–geometric means inequality, we have
sin 2 A 2 + sin2 B 2 + sin2 C 2 ≥ 3t2 .
By Problem 22, we have 3t 2 + 2t 3 ≤ 1. Thus,
0 ≥ 2t 3 + 3t 2 − 1 = (t + 1)(2t 2 + t − 1) = (t + 1) 2 (2t − 1).
Consequently, t ≤
2 1 , establishing (a).
24. In triangle ABC, show that
(a) sin 2A + sin 2B + sin 2C = 4 sin A sin B sin C;
(b) cos 2A + cos 2B + cos 2C =−1 − 4 cos A cos B cos C;
(c) sin 2 A + sin 2 B + sin 2 C = 2 + 2 cos A cos B cos C;
(d) cos 2 A + cos 2 B + cos 2 C + 2 cos A cos B cos C = 1.
Conversely, if x,y,z are positive real numbers such that
x 2 + y 2 + z 2 + 2xyz = 1,
show that there is an acute triangle ABC such that x = cos A, y = cos B,
C = cos C.