103 Trigonometry Problems

168 103 Trigonometry Problems
Applying the law of sines to triangle A 1 A 2 A 3 , the last inequality reduces to
3∑
sin A i (cos A i − 2 cos A i+1 A i+2 ) ≥ 0,
i=1
which follows directly from the next Lemma.
Lemma
Proof:
Let ABC be a triangle. Then the Cyclic Sum
∑
sin A(cos A − 2 cos B cos C) = 0.
cyc
By the double-angle formulas, it suffices to show that
∑
sin 2A = 2 ∑ sin A cos A = 4 ∑ sin A cos B cos C.
cyc
cyc
cyc
Applying the addition and subtraction formulas gives
sin A cos B cos C + sin B cos C cos A
= cos C(sin A cos B + sin B cos A)
= cos C sin(A + B) = cos C sin C.
Hence
4 ∑ sin A cos B cos C
cyc
= 2 ∑ cyc
(sin A cos B cos C + sin B cos C cos A)
= 2 ∑ cos C sin C = ∑ sin 2C,
cyc
cyc
as desired.
42. Let ABC be a triangle. Let x,y, and z be real numbers, and let n be a positive
integer. Prove the following four inequalities.
(a) [D. Barrow] x 2 + y 2 + z 2 ≥ 2yzcos A + 2zx cos B + 2xy cos C.
(b) [J. Wolstenholme]
x 2 + y 2 + z 2 ≥ 2(−1) n+1 (yz cos nA + zx cos nB + xy cos nC).