103 Trigonometry Problems

130 103 Trigonometry Problems
7. Let ABC be an acute-angled triangle with area K. Prove that
√
√
a 2 b 2 − 4K 2 + b 2 c 2 − 4K 2 +
√c 2 a 2 − 4K 2 = a2 + b 2 + c 2
.
2
Solution: We have 2K = ab sin C = bc sin A = ca sin B. The expression on
the left-hand side of the desired equation is equal to
√
a 2 b 2 − a 2 b 2 sin 2 C +
√
b 2 c 2 − b 2 c 2 sin 2 A +
= ab cos C + bc cos A + ca cos B
= a 2 (b cos C + c cos B) + b (c cos A + a cos C)
2
+ c (a cos B + b cos A)
2
= a 2 · a + b 2 · b + c 2 · c,
√
c 2 a 2 − c 2 a 2 sin 2 B
and the conclusion follows.
Note: We encourage the reader to explain why this problem is the equality
case of Advanced Problem 42(a).
8. Compute the sums
( ( ( n n n
sin a + sin 2a +···+ sin na
1)
2)
n)
and ( ( ( n n n
cos a + cos 2a +···+ cos na.
1)
2)
n)
Solution: Let S n and T n denote the first and second sums, respectively. Set
the complex number z = cos a + i sin a. Then, by de Moivre’s formula, we
have z n = cos na + i sin na.Bythebinomial theorem, we obtain
( ( n n
1 + T n + iS n = 1 + (cos a + i sin a) + (cos 2a + i sin 2a)
1)
2)
( n
+···+ (cos na + i sin na)
n)
=
( n
0)
z 0 +
= (1 + z) n .
( n
1)
z +
( n
2)
z 2 +···+
( n
n)
z n