103 Trigonometry Problems

4. Solutions to Introductory Problems 115
47. Prove that
| sin a 1 |+|sin a 2 |+···+|sin a n |+|cos(a 1 + a 2 +···+a n )|≥1.
Solution: We proceed by induction on n. The base case holds, because
| sin a 1 |+|cos a 1 |≥sin 2 a 1 + cos 2 a 1 = 1.
For the inductive step, in order to prove that
| sin a 1 |+|sin a 2 |+···+|sin a n+1 |+|cos(a 1 + a 2 +···+a n+1 )|≥1,
it suffices to show that
| sin a n+1 |+|cos(a 1 + a 2 +···+a n+1 )|≥|cos(a 1 + a 2 +···+a n )|
for all real numbers a 1 ,a 2 ,...,a n+1 . Let s k = a 1 + a 2 +···+a k , for k =
1, 2,...,n+1. The last inequality becomes | sin a n+1 |+| cos s n+1 |≥|cos s n |.
Indeed, by the addition and subtraction formulas, wehave
as desired.
| cos s n |=|cos(s n+1 − a n+1 )|
=|cos s n+1 cos a n+1 + sin s n+1 sin a n+1 |
=|cos s n+1 cos a n+1 |+|sin s n+1 sin a n+1 |
≤|cos s n+1 |+|sin a n+1 |,
48. [Russia 2003, by Nazar Agakhanov] Find all angles α for which the threeelement
set
S ={sin α, sin 2α, sin 3α}
is equal to the set
T ={cos α, cos 2α, cos 3α}.
Solution: The answers are α = π 8 + kπ 2
for all integers k.
Because S = T , the sums of the elements in S and T are equal to each other;
that is,
sin α + sin 2α + sin 3α = cos α + cos 2α + cos 3α.
Applying the sum-to-product formulas to the first and the third summands
on each side of the last equation gives
2 sin 2α cos α + sin 2α = 2 cos 2α cos α + cos 2α,