103 Trigonometry Problems

126 103 Trigonometry Problems
(a) Multiplying both sides of the given equation by sin 1 ◦ ,wehave
sin 1 ◦
sin n ◦ = (cot 45◦ − cot 46 ◦ ) + (cot 47 ◦ − cot 48 ◦ )
+···+(cot 133 ◦ − cot 134 ◦ )
= cot 45 ◦ − (cot 46 ◦ + cot 134 ◦ ) + (cot 47 ◦ + cot 133 ◦ )
−···+(cot 89 ◦ + cot 91 ◦ ) − cot 90 ◦
= 1.
Therefore, sin n ◦ = sin 1 ◦ , and the least possible integer value for n is 1.
(b) The left-hand side of the desired equation is equal to
89
∑
k=1
1
sin k ◦ sin(k + 1) ◦ = 1 ∑89
sin 1 ◦
thus completing the proof.
k=1
[
cot k ◦ − cot(k + 1) ◦]
= 1
sin 1 ◦ · cot cos 1◦ 1◦
=
sin 2 1 ◦ ,
2. [China 2001, by Xiaoyang Su] Let ABC be a triangle, and let x be a nonnegative
real number. Prove that
a x cos A + b x cos B + c x cos C ≤ 1 2 (ax + b x + c x ).
Solution: By symmetry, we may assume that a ≥ b ≥ c. Hence A ≥ B ≥ C,
and so cos A ≤ cos B ≤ cos C. Thus
or
(a x − b x )(cos A − cos B) ≤ 0,
a x cos A + b x cos B ≤ a x cos B + b x cos A.
Adding the last inequality with its analogous cyclic symmetric forms and then
adding a x cos A + b x cos B + c x cos C to both sides of the resulting inequality
gives
3(a x cos A + b x cos B + c x cos C)
≤ (a x + b x + c x )(cos A + cos B + cos C),