103 Trigonometry Problems

4. Solutions to Introductory Problems 91
Solution: The equality is equivalent to
or
tan 3a(1 − tan 2a tan a) = tan 2a + tan a,
tan 3a =
tan 2a + tan a
1 − tan 2a tan a .
That is, tan 3a = tan(2a + a), which is evident.
Note: More generally, if a 1 ,a 2 ,a 3 are real numbers different from kπ 2 , where
k is in Z, such that a 1 + a 2 + a 3 = 0, then the relation
tan a 1 + tan a 2 + tan a 3 = tan a 1 tan a 2 tan a 3
holds. The proof of this relation is similar to the proofs of Problems 13 and
20. We leave the proof as an exercise for the reader.
14. Let a, b, c, d be numbers in the interval [0,π] such that
sin a + 7 sin b = 4(sin c + 2 sin d),
cos a + 7 cos b = 4(cos c + 2 cos d).
Prove that 2 cos(a − d) = 7 cos(b − c).
Solution: Rewrite the two given equalities as
sin a − 8 sin d = 4 sin c − 7 sin b,
cos a − 8 cos d = 4 cos c − 7 cos b.
By squaring the last two equalities and adding them, we obtain
1+64−16(cos a cos d +sin a sin d) = 16+49−56(cos b cos c+sin b sin c),
and the conclusion follows from the addition formulas.
15. Express
sin(x − y) + sin(y − z) + sin(z − x)
as a monomial.