103 Trigonometry Problems

60 103 Trigonometry Problems
then r = √ a 2 + b 2 and a = r cos θ and b = r sin θ. We write z = a + bi =
r cos θ + ri sin θ = r cis θ. For example, z =−1 + √ 3i = 2 cis 120 ◦ = 2 cis 2π 3 .
We have developed interesting properties of adding vectors with equal magnitudes.
Similar properties can be written in terms of complex numbers. Let z 1 and z 2 be
complex numbers with |z 1 |=|z 2 |=r. Set z 1 = r cis θ 1 and z 2 = r cis θ 2 . Set
z = z 1 + z 2 . Then OZ 1 ZZ 2 is a rhombus, implying that the line OZ bisects the
angle Z 1 OZ 2 ; that is, z = r ′ cis θ 1+θ 2
2
for some real number r ′ . In particular, if
z 1 = 1 and z 2 = cis a, then z = r ′ cis a 2 , where r′ = OZ. Therefore, tan a 2
is equal
to the slope of line OZ; that is,
tan a 2 =
sin a
1 + cos a ,
which is one of half-angle formulas. Other versions of the half-angle formulas can
be obtained in a similar fashion. It is also not difficult to see that r ′ = OZ = 2 cos a 2 .
Example 1.23. [AMC12 2002] Let a and b be real numbers such that
Evaluate sin(a + b).
√
2
sin a + sin b =
2 ,
√
6
cos a + cos b =
2 .
First Solution: Square both of the given relations and add the results to obtain
sin 2 a + cos 2 a + sin 2 b + cos 2 b + 2(sin a sin b + cos a cos b) = 2 4 + 6 4 ,
or cos(a − b) = 2(sin a sin b + cos a cos b) = 0 by the addition and subtraction
formulas. Multiplying the two given relations yields
√
3
sin a cos a + sin b cos b + sin a cos b + sin b cos a =
2 ,
or sin 2a + sin 2b + 2 sin(a + b) = √ 3bythedouble-angle formulas and the
addition and subtraction formulas. By the sum-to-product formulas, we obtain
Hence sin(a + b) =
sin 2a + sin 2b = 2 sin(a + b) cos(a − b) = 0.
√
3
2 .
Second Solution:
Set complex numbers z 1 = cos a +i sin a = cis a and z 2 = cos b +i sin b = cis b
(Figure 1.59).