103 Trigonometry Problems

186 103 Trigonometry Problems
or
7 ≥ tan 2 θ 2 + tan 2 θ 3 .
Thus if tan 2 θ 2 + tan 2 θ 3 ≤ 7, then inequality (∗) holds, and we are done.
Assume that tan 2 θ 2 + tan 2 θ 3 > 7. Then tan 2 θ 1 ≥ tan 2 θ 2 ≥
2 7 . Then
√
1 2
cos θ 1 ≤ cos θ 2 = √ ≤
1 + tan 2 θ 2
3 ,
implying that
establishing (∗) again.
cos θ 1 + cos θ 2 + cos θ 3 ≤ 2√ 2
3
Therefore, inequality (∗) is true, as desired.
+ 1 < 2,
48. Let ABC be an acute triangle. Prove that
(sin 2B + sin 2C) 2 sin A + (sin 2C + sin 2A) 2 sin B
+ (sin 2A + sin 2B) 2 sin C ≤ 12 sin A sin B sin C.
First Solution: Applying the addition and subtraction formulas gives
(sin 2B + sin 2C) 2 sin A = 4 sin 2 (B + C)cos 2 (B − C)sin A
= 4 sin 3 A cos 2 (B − C),
because A + B + C = 180 ◦ . Hence it suffices to show that the cyclic sum
∑
sin 3 A cos 2 (B − C)
cyc
is less than or equal to 3 sin A sin B sin C, which follows from
∑
4 sin 3 A cos(B − C) = 12 sin A sin B sin C.
cyc
Indeed, we have
4 sin 3 A cos(B − C)
= 4 sin 2 A sin(B + C)cos(B − C)
= 2 sin 2 A(sin 2B + sin 2C)
= (1 − cos 2A)(sin 2B + sin 2C)
= (sin 2B + sin 2C) − sin 2B cos 2A − sin 2C cos 2A.