103 Trigonometry Problems

1. Trigonometric Fundamentals 25
C
C
D
D
E
B
A
B
Figure 1.24.
A
Second Solution: Note that ̸ CDA = 60 ◦ and that sin 30 ◦ =
2 1 . We construct point
E on segment AD (Figure 1.24, right) such that CE ⊥ AD. Then in triangle CDE,
̸ DCE = 30 ◦ and |DE| =|CD| sin ̸ DCE, or|CD| =2|DE|. Thus triangle
BDE is isosceles with |DE| =|DB|, implying that ̸ DBE = ̸ DEB = 30 ◦ .
Consequently, ̸ CBE = ̸ BCE = 30 ◦ and ̸ EBA = ̸ EAB = 15 ◦ , and so
triangles BCE and BAE are both isosceles with |CE| =|BE| =|EA|. Hence
the right triangle AEC is isosceles; that is, ̸ ACE = ̸ EAC = 45 ◦ . Therefore,
̸ ACB = ̸ ACE + ̸ ECB = 75 ◦ .
For a function f : A → B, iff (A) = B, then f is said to be surjective (or
onto); that is, every b ∈ B is the image under f of some a ∈ A. If every two distinct
elements a 1 and a 2 in A have distinct images, then f is injective (or one-to-one). If
f is both injective and surjective, then f is bijective (or a bijection or a one-to-one
correspondence).
The sine and cosine functions are functions from the set of angles to the real
numbers. The images of the two functions are the real numbers between −1 and 1.
For a point P = (x, y) with polar coordinates (1,θ) on the unit circle, it is clear
that the values x = cos θ and y = sin θ vary continuously from −1 to 1, taking
on all intermediate values. Hence the functions are surjective functions from the
set of angles to the interval [−1, 1]. On the other hand, these two functions are not
one-to-one. It is not difficult to see that the sine function is a bijection between the
set of angles α with −90 ◦ ≤ α ≤ 90 ◦ and the interval [−1, 1], and that the cosine
function is a bijection between the set of angles α with 0 ◦ ≤ α ≤ 180 ◦ and the
interval [−1, 1]. For abbreviation, we can write that sin :[−90 ◦ , 90 ◦ ]→[−1, 1] is
a bijection. It is also not difficult to see that the tangent function is a bijection between
the set of angles α with −90 ◦