Download 12 Trigonometric Functions of Any Angle

Arkansas Tech University
MATH 1203: Trigonometry
Dr. Marcel B. Finan
12
Trigonometric Functions of Any Angle
Right triangles are useful when trying to calculate the trigonometric functions of acute angles. What about angles that are not acute angles?
In this section you will learn (1) of how to compute the trigonometric functions of any angle, not just acute angles, (2) the sign of the trigonometric
functions in each quadrant of the coordinate plane, and (3) the use of reference angles which reduce the question of finding the trigonometric functions
of an angle to that of finding the trigonometric functions of the special angles 30◦ , 45◦ , and 60◦ .
Let θ be an angle in standard position as shown in Figure 12.1.
Figure 12.1
Let P (x, y) be any point on the terminal side. If r is the distance
p from the
origin to the point P then by the Pythagorean Theorem, r = x2 + y 2 . We
define the trigonometric functions of θ to be
sin θ =
csc θ =
y
r
r
y
cos θ =
sec θ =
x
r
r
x
tan θ =
cot θ =
y
x
x
y
where x 6= 0 and y 6= 0. If θ = kπ, where k is an integer, then the functions
csc θ and cot θ are undefined since a point on the terminal side has components P (x, 0). Similarly, if θ = (2k + 1) π2 then the functions sec θ and tan θ
are undefined since P (0, y).
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Question: Do the above definitions depend on the choice of the point P ?
The answer is no. To see this, let Q(x0 , y 0 ) be any other point on the terminal side of θ. See Figure 12.2.
Figure 12.2
Then the right triangles ∆OP M and ∆OQN are similar triangles since
corresponding angles are equal. This implies that the ratios of the corresponding sides are equal. Thus
x
y
r
= 0 = 0
0
r
x
y
or equivalently
x
x0
y
y0
y
y0
= 0 = = 0 = = 0
r
r
r
r
x
x
Thus,
sin θ =
csc θ =
y0
r00
r
y0
cos θ =
sec θ =
x0
r0
r0
x0
tan θ =
cot θ =
y0
x0 0
x
y0 .
This shows that the trigonometric functions are independent of the point
chosen on the terminal side of the angle.
Example 12.1
Complete the following chart, using the definitions introduced above.
θ
0◦
90◦
180◦
270◦
sin θ
cos θ
tan θ
2
csc θ
sec θ
cot θ
Solution.
For θ = 0◦ we choose the point P (1, 0). For θ = 90◦ we choose P (0, 1); for
θ = 180◦ , we choose P (−1, 0) and finally for θ = 270◦ we choose P (0, −1).
Then by the above definitions we have
θ
0◦
90◦
180◦
270◦
sin θ
0
1
0
-1
cos θ
1
0
-1
0
tan θ
0
undefined
0
undefined
csc θ
undefined
1
undefined
-1
sec θ
1
undefined
-1
undefined
cot θ
undefined
0
undefined
0
Example 12.2
Find the exact value of the six trigonometric functions of an angle θ if
P (4, −3) is a point on its terminal side.
Solution.
√
First, note that r = 16 + 9 = 5. Thus,
4
−3
sin θ = −3
5 ; cos θ = 5 ; tan θ = 4
5
−4
csc θ = −5
3 ; sec θ = 4 ; cot θ = 3 .
Example 12.3
Given that cos θ = − 32 , π2 < θ < π, find the exact value of each of the
remaining trigonometric functions.
Solution.
Since θ is in Quadrant II, sin θ > 0, csc θ > 0, sec θ < 0, tan θ < √
0, cot θ < 0.
Thus, x = −2, r = 3 and by the Pythagorean
formula
y
=
r 2 − x2 =
√
√
√
√
√
9 − 4 = 5. It follows that sin θ = 35 . So csc θ = 3 5 5 , tan θ = − 25 , cot θ =
√
− 2 5 5 , and finally sec θ = − 32 .
Example 12.4
Complete the following chart of signs of the six trigonometric functions.
Q
I
II
III
IV
sin x
cos x
tan x
Solution.
3
cot x
sec x
csc x
Q
I
II
III
IV
sin x
+
+
−
−
cos x
+
−
−
+
tan x
+
−
+
−
cot x
+
−
+
−
sec x
+
−
−
+
csc x
+
+
−
−
Reference angles
The values of trigonometric functions of angles greater than 90◦ (or less
than 0◦ ) can be determined from their values at corresponding angles called
reference angles.
For an angle θ in standard position, the acute angle θ0 between the terminal
side of θ and either the positive or negative x-axis is called the reference
angle of θ. Figure 12.3 illustrates the reference angles for some general
angles θ.
Figure 12.3
Example 12.5
Determine the reference angles for the following given angles:
(a) −70◦ (b) 255◦ (c) 5π
3 rad
Solution.
(a) θ0 = 70◦ (b) θ0 = 75◦ (c) θ0 = π3 .
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Remark 12.1
Note that if θ is a nonacute angle and θ0 is its reference angle then the
trigonometric values of θ are equal to the trigonometric values of θ0 with the
appropriate sign which depends on the quadrant in which θ lies.
Example 12.6
Evaluate the following:
11π
◦
(a) cos 4π
3 (b) tan (−210 ) (c) csc 3 .
Solution.
π
1
(a) cos 4π
3 = − cos 3 = − 2 .
√
(b) tan (−210◦ ) = − tan 210◦ = − tan π6 = − 33 .
√
3π
3π
π
(c) csc 11π
2.
3 = csc (2π + 4 ) = csc 4 = csc 4 =
Example 12.7
Referring to Figure 12.4, answer the following questions.
(a) Express the area of ∆OBC in terms of sin θ and cos θ.
(b) Express the area of ∆OBD in terms of tan θ.
(c) Use parts (a) and (b) to show
1<
θ
1
<
.
sin θ
cos θ
Figure 12.4
Solution.
(a) Area ∆OBC = 21 |EC||OB| = 12 |EC| =
|EC|.
(b) Area ∆OBD = 21 |BD||OB| = 12 |BD| =
|DB|.
(c) Using Figure 12.4 we see that
1
2
sin θ since sin θ =
|EC|
|OC|
=
1
2
tan θ since tan θ =
|DB|
|OB|
=
Area ∆OBC < Area circular sector OBC < Area ∆OBD.
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But the area of the circular sector OBC is 21 θ. Hence,
1
1
1
sin θ < θ < tan θ.
2
2
2
Multiplying through by
2
sin θ
to obtain
1<
θ
1
<
.
sin θ
cos θ
Note that according to the given figure the angle is assumed to be in the
first quadrant so that the sine function is positive there.
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