103 Trigonometry Problems

138 103 Trigonometry Problems
or cos a + cos(b + e) = cos e + cos(a + b). Applying the sum-to-product
formulas again gives
2 cos a + b + e
2
cos a − b − e
2
= 2 cos a + b + e
2
cos e − a − b ,
2
and so cos a−b−e
2
= cos e−a−b
2
. It follows that a = e. Because a = e and
b = d, triangles ABC and AED are congruent.
16. All the angles in triangle ABC are less then 120 ◦ . Prove that
cos A + cos B − cos C
sin A + sin B − sin C > − √
3
3 .
Solution: Consider the triangle A 1 B 1 C 1 , as shown in Figure 5.3, where ̸ A 1 =
120 ◦ − ̸ A, ̸ B 1 = 120 ◦ − ̸ B, and ̸ C 1 = 120 ◦ − ̸ C. The given condition
guarantees the existence of such a triangle.
C
A1
C1
A
B
B1
Figure 5.3.
Applying the triangle inequality in triangle A 1 B 1 C 1 gives B 1 C 1 + C 1 A 1 >
A 1 B 1 ; that is
sin A 1 + sin B 1 > sin C 1
by applying the law of sines to triangle A 1 B 1 C 1 . It follows that
sin(120 ◦ − A) + sin(120 ◦ − B) > sin(120 ◦ − C),
or √
3
2 (cos A + cos B − cos C) + 1 (sin A + sin B − sin C) > 0.
2
Taking into account that a + b>cimplies sin A + sin B − sin C>0, the
above inequality can be rewritten as
√
3 cos A + cos B − cos C
·
2 sin A + sin B − sin C + 1 2 > 0,
from which the conclusion follows.