103 Trigonometry Problems

122 103 Trigonometry Problems
By Gauss’s lemma, the only possible rational roots of the above cubic
equation are 1 and −1. It is easy to see that neither is a root. Hence the
above equation has no rational root, implying that 2 cos a is not rational.
Therefore, cos a is not rational.
Note: Although converting to equation (∗) is not necessary, it is a very
effective technique. Instead of checking of eight possible rational roots
from the set { ± 1 8 , ± 1 4 , ± 1 2 , ±1 } of the equation
8x 3 − 4x 2 − 4x + 1 = 0,
we need to check only two possibilities for equation (∗).
(f) Because 3a + 4a = π, it follows that tan 3a + tan 4a = 0. The doubleangle
and the addition and subtraction formulas yield
or
tan a + tan 2a 2 tan 2a
+
1 − tan a tan 2a 1 − tan 2 2a = 0,
tan a + 3 tan 2a − 3 tan a tan 2 2a − tan 3 2a = 0.
Set tan a = x. Then tan 2a =
2 tan a
1−tan 2 a =
2x
1−x 2 . Hence
x +
6x
1 − x 2 − 12x3
(1 − x 2 ) 2 − 8x3
(1 − x 2 ) 3 = 0,
or (1 − x 2) 3 (
+ 6 1 − x 2) 2 ( − 12x
2
1 − x 2) − 8x 2 = 0.
Expanding the left-hand side of the above equation gives
x 6 − 21x 4 + 35x 2 − 7 = 0. (†)
Thus tan a is a root of the above equation. Note that 6a + 8a = 2π and
9a + 12a = 3π, and so tan[3(2a)]+tan[4(2a)] =0 and tan[3(3a)]+
tan[4(3a)] =0. Hence tan 2a and tan 3a are the also the roots of equation
(†). Therefore, tan 2 ka, k = 1, 2, 3, are the distinct roots of the cubic
equation
x 3 − 21x 2 + 35x − 7 = 0.
By Viète’s theorem,wehave
tan 2 a + tan 2 2a + tan 2 3a = 21;
tan 2 a tan 2 2a + tan 2 2a tan 2 3a + tan 2 3a tan 2 a = 35;
tan 2 a tan 2 2a tan 2 3a = 7.