103 Trigonometry Problems

4. Solutions to Introductory Problems 121
(a) By the sum-to-product, the difference-to-product and the doubleangle
formulas,wehave
sin 2 3a − sin 2 a = (sin 3a + sin a)(sin 3a − sin a)
= (2 sin 2a cos a)(2 sin a cos 2a)
= (2 sin 2a cos 2a)(2 sin a cos a)
= sin 4a sin 2a = sin 2a sin 3a,
as desired. The last identity is evident by noting that 4a + 3a = π (and
so sin 3a = sin 4a).
(b) It suffices to show that
or
sin 2a sin 4a = sin a(sin 2a + sin 4a),
2 sin a cos a sin 4a = sin a(2 sin 3a cos a),
by the sum-to-product formulas.
(c) The answer is 1 2 . It suffices to show that cos 2a + cos 4a + cos 6a =−1 2 .
This is a special case (n = 3) of a more general result:
t = cos 2x + cos 4x +···+cos 2nx =− 1 2 ,
π
where x =
2n+1
. Indeed, applying the product-to-sum formulas gives
2 sin x cos kx = sin(k + 1)x − sin(k − 1)x, and so
2t sin x = 2 sin x(cos 2x + cos 4x +···+cos 2nx)
=[sin 3x − sin x]+[sin 5x − sin 3x]
+···+[sin(2n + 1)x − sin(2n − 1)x]
= sin(2n + 1)x − sin x =−sin x,
from which the desired equality follows.
(d) Because 3a + 4a = π, it follows that sin 3a = sin 4a. The double-angle
and triple-angle formulas yield
sin a(3 − 4 sin 2 a) = 2 sin 2a cos 2a = 4 sin a cos a cos 2a,
or 3 − 4(1 − cos 2 a) = 4 cos a(2 cos 2 a − 1). It follows that
8 cos 3 a − 4 cos 2 a − 4 cos a + 1 = 0,
establishing (c). Thus u = 2 cos a is the root of the cubic equation
u 3 − u 2 − 2u + 1 = 0.
(∗)