103 Trigonometry Problems

4. Solutions to Introductory Problems 99
of the system is 0; that is,
−1 cos B cos C
0 =
cos B −1 cos A
∣ cos C cos A −1 ∣
as desired.
=−1 + 2 cos A cos B cos C + cos 2 A + cos 2 B + cos 2 C,
25. In triangle ABC, show that
(a) 4R =
abc
[ABC] ;
(b) 2R 2 sin A sin B sin C =[ABC];
(c) 2R sin A sin B sin C = r(sin A + sin B + sin C);
(d) r = 4R sin A 2 sin B 2 sin C 2 ;
(e) a cos A + b cos B + c cos C = abc
2R 2 .
Solution: By the extended law of sines,
R =
a
2 sin A = abc
2bc sin A = abc
4[ABC] ,
establishing (a). By the same token, we have
which is (b).
Note that
2R 2 sin A sin B sin C = 1 · (2R sin A)(2R sin B)(sin C)
2
= 1 ab sin C =[ABC],
2
2[ABC] =bc sin A = (a + b + c)r.
By the extended law of sines, we obtain
4R 2 sin A sin B sin C = bc sin A = r(a + b + c)
= 2rR(sin A + sin B + sin C),
from which (c) follows.