103 Trigonometry Problems

4. Solutions to Introductory Problems 101
26. Let s be the semiperimeter of triangle ABC. Prove that
(a) s = 4R cos A 2 cos B 2 cos C 2 ;
(b) s ≤ 3√ 3
2 R.
Solution: It is well known that rs =[ABC], ors = [ABC]
r
. By Problem 25
(b) and (d), part (a) follows from
R sin A sin B sin C
s =
2 sin A 2 sin B 2 sin C 2
= 4R cos A 2 cos B 2 cos C 2
by the double-angle formulas.
We conclude part (b) from (a) and Problem 23 (d).
27. In triangle ABC, show that
(a) cos A + cos B + cos C = 1 + 4 sin A 2 sin B 2 sin C 2 ;
(b) cos A + cos B + cos C ≤ 3 2 .
Solution: By the sum-to-product and the double-angle formulas, wehave
cos A + cos B = 2 cos A + B
2
cos A − B
2
= 2 sin C 2 cos A − B
2
and
1 − cos C = 2 sin 2 C 2 = 2 sin C 2 cos A + B .
2
It suffices to show that
2 sin C [
cos A − B − cos A + B ]
= 4 sin A 2 2
2
2 sin B 2 sin C 2 ,
or,
cos A − B − cos A + B = 2 sin A 2
2 2 sin B 2 ,
which follows from the sum-to-product formulas, and hence (a) is established.
Recalling Problem 25 (c), we have
cos A + cos B + cos C = 1 + r R .
(∗)