103 Trigonometry Problems

92 103 Trigonometry Problems
Solution: By the sum-to-product formulas, wehave
sin(x − y) + sin(y − z) = 2 sin x − z
2
By the double-angle formulas,wehave
Thus,
sin(z − x) = 2 sin z − x
2
sin(x − y) + sin(y − z) + sin(z − x)
= 2 sin x − z
2
=−4 sin x − z
2
=−4 sin x − y
2
by the sum-to-product formulas.
cos x + z − 2y .
2
cos z − x
2 .
[
cos x + z − 2y − cos z − x
2
2
sin z − y
2
sin y − z
2
sin x − y
2
sin z − x
2
Note: In exactly the same way, we can show that if a, b, and c are real numbers
with a + b + c = 0, then
sin a + sin b + sin c =−4 sin a 2 sin b 2 sin c 2 .
In Problem 15, we have a = x − y, b = y − z, and c = z − x.
]
16. Prove that
(4 cos 2 9 ◦ − 3)(4 cos 2 27 ◦ − 3) = tan 9 ◦ .
Solution: We have cos 3x = 4 cos 3 x − 3 cos x, so4cos 2 x − 3 =
all x ̸= (2k + 1) · 90 ◦ , k ∈ Z. Thus
cos 3x
cos x
for
(4 cos 2 9 ◦ − 3)(4 cos 2 27 ◦ cos 27◦ cos 81◦ cos 81◦
− 3) = · =
cos 9◦ cos 27◦ cos 9 ◦
sin 9◦
=
cos 9 ◦ = tan 9◦ ,
as desired.