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Copyright © Cengage Learning. All rights reserved.
4.4 Trigonometric Functions
of Any Angle
2
What You Should Learn
• Evaluate trigonometric functions of any angle
• Find reference angles
• Evaluate trigonometric functions of real
numbers
3
Introduction
4
Introduction
Following is the definition of trigonometric functions of Any
Angle. This applies when the radius is not one (not a unit
circle).
5
Introduction
Note: when x = 0, the tangent and secant of  are
undefined.
For example, the tangent of 90 is undefined since the sine
of 90 is 1 and the cosine of 90 is 0. 1/0 is undefined.
Similarly, when y = 0, the cotangent and cosecant of  are
undefined.
6
Example 1 – Evaluating Trigonometric Functions
Let (–3, 4) be a point on the terminal side of 
(see Figure 4.34).
Find the sine, cosine, and tangent of .
Figure 4.34
7
Example 1 – Solution
Referring to Figure 4.34, you can see by using the
Pythagorean Theorem and the given point that x = –3, y =
4, and
8
Example 1 – Solution
So, you have
and
cont’d
9
Introduction
The signs of the trigonometric functions in the four
quadrants can be determined easily from the definitions of
the functions. For instance, because
it follows that cos  is positive
wherever x > 0, which is in
Quadrants I and IV.
We will discuss “All Students
Take Calculus” in class as a way
to help us remember this.
10
Reference Angles
11
Reference Angles
The values of the trigonometric functions of angles greater
than 90 (or less than 0) can be determined from their
values at corresponding acute angles called reference
angles.
12
Reference Angles
Figure 4.37 shows the reference angles for  in Quadrants
II, III, and IV.
Figure 4.37
13
Example 4 – Finding Reference Angles
Find the reference angle .
a.  = 300 b.  = 2.3 c.  = –135
Solution:
a. Because 300 lies in Quadrant IV, the angle it makes
with the x-axis is
 = 360 – 300
= 60.
Degrees
14
Example 4 – Solution
b. Because 2.3 lies between /2  1.5708 and   3.1416,
it follows that it is in Quadrant II and its reference angle
is
 =  – 2.3
 0.8416.
c. First, determine that –135 is coterminal with 225,
which lies in Quadrant III. So, the reference angle is
 = 225 – 180
= 45.
cont’d
Radians
Degrees
15
Example 4 – Solution
Figure 4.38 shows each angle  and its reference angle .
cont’d
(a) (b) (c)
Figure 4.38
16
Trigonometric Functions of Real Numbers
17
Trigonometric Functions of Real Numbers
To see how a reference angle is used to evaluate a
trigonometric function, consider the point (x, y) on the
terminal side of , as shown in Figure 4.39.
Figure 4.39
18
Trigonometric Functions of Real Numbers
By definition, you know that
and
For the right triangle with acute angle  and sides of
lengths |x| and |y|, you have
and
19
Trigonometric Functions of Real Numbers
So, it follows that sin  and sin  are equal, except
possibly in sign. The same is true for tan  and tan  and
for the other four trigonometric functions. In all cases, the
sign of the function value can be determined by the
quadrant in which  lies.
20
Example 5 – Trigonometric Functions of Nonacute Angles
Evaluate each trigonometric function.
a. b. tan = (–210) c.
Solution:
a. Because  = 4/3 lies in Quadrant III, the
reference angle is   = (4/3) –  = /3,
as shown in Figure 4.40.
Moreover, the cosine is negative in
Quadrant III, so
Figure 4.40
21
Example 5 – Solution
b. Because –210 + 360 = 150, it follows that –210 is
coterminal with the second-quadrant angle 150.
Therefore, the reference angle is  = 180 – 150 = 30,
as shown in Figure 4.41.
Finally, because the tangent is
negative in Quadrant II, you have.
cont’d
Figure 4.41
22
Example 5 – Solution
c. Because (11/4) – 2 = 3/4, it follows that 11/4 is
coterminal with the second-quadrant angle 3/4.
Therefore, the reference angle is  =  – (3/4) = /4,
as shown in Figure 4.42.
Because the cosecant is positive
in Quadrant II, you have
cont’d
Figure 4.42

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trigonometric-functions-of-any-angle.pdf

  • 1. Copyright © Cengage Learning. All rights reserved. 4.4 Trigonometric Functions of Any Angle
  • 2. 2 What You Should Learn • Evaluate trigonometric functions of any angle • Find reference angles • Evaluate trigonometric functions of real numbers
  • 4. 4 Introduction Following is the definition of trigonometric functions of Any Angle. This applies when the radius is not one (not a unit circle).
  • 5. 5 Introduction Note: when x = 0, the tangent and secant of  are undefined. For example, the tangent of 90 is undefined since the sine of 90 is 1 and the cosine of 90 is 0. 1/0 is undefined. Similarly, when y = 0, the cotangent and cosecant of  are undefined.
  • 6. 6 Example 1 – Evaluating Trigonometric Functions Let (–3, 4) be a point on the terminal side of  (see Figure 4.34). Find the sine, cosine, and tangent of . Figure 4.34
  • 7. 7 Example 1 – Solution Referring to Figure 4.34, you can see by using the Pythagorean Theorem and the given point that x = –3, y = 4, and
  • 8. 8 Example 1 – Solution So, you have and cont’d
  • 9. 9 Introduction The signs of the trigonometric functions in the four quadrants can be determined easily from the definitions of the functions. For instance, because it follows that cos  is positive wherever x > 0, which is in Quadrants I and IV. We will discuss “All Students Take Calculus” in class as a way to help us remember this.
  • 11. 11 Reference Angles The values of the trigonometric functions of angles greater than 90 (or less than 0) can be determined from their values at corresponding acute angles called reference angles.
  • 12. 12 Reference Angles Figure 4.37 shows the reference angles for  in Quadrants II, III, and IV. Figure 4.37
  • 13. 13 Example 4 – Finding Reference Angles Find the reference angle . a.  = 300 b.  = 2.3 c.  = –135 Solution: a. Because 300 lies in Quadrant IV, the angle it makes with the x-axis is  = 360 – 300 = 60. Degrees
  • 14. 14 Example 4 – Solution b. Because 2.3 lies between /2  1.5708 and   3.1416, it follows that it is in Quadrant II and its reference angle is  =  – 2.3  0.8416. c. First, determine that –135 is coterminal with 225, which lies in Quadrant III. So, the reference angle is  = 225 – 180 = 45. cont’d Radians Degrees
  • 15. 15 Example 4 – Solution Figure 4.38 shows each angle  and its reference angle . cont’d (a) (b) (c) Figure 4.38
  • 17. 17 Trigonometric Functions of Real Numbers To see how a reference angle is used to evaluate a trigonometric function, consider the point (x, y) on the terminal side of , as shown in Figure 4.39. Figure 4.39
  • 18. 18 Trigonometric Functions of Real Numbers By definition, you know that and For the right triangle with acute angle  and sides of lengths |x| and |y|, you have and
  • 19. 19 Trigonometric Functions of Real Numbers So, it follows that sin  and sin  are equal, except possibly in sign. The same is true for tan  and tan  and for the other four trigonometric functions. In all cases, the sign of the function value can be determined by the quadrant in which  lies.
  • 20. 20 Example 5 – Trigonometric Functions of Nonacute Angles Evaluate each trigonometric function. a. b. tan = (–210) c. Solution: a. Because  = 4/3 lies in Quadrant III, the reference angle is   = (4/3) –  = /3, as shown in Figure 4.40. Moreover, the cosine is negative in Quadrant III, so Figure 4.40
  • 21. 21 Example 5 – Solution b. Because –210 + 360 = 150, it follows that –210 is coterminal with the second-quadrant angle 150. Therefore, the reference angle is  = 180 – 150 = 30, as shown in Figure 4.41. Finally, because the tangent is negative in Quadrant II, you have. cont’d Figure 4.41
  • 22. 22 Example 5 – Solution c. Because (11/4) – 2 = 3/4, it follows that 11/4 is coterminal with the second-quadrant angle 3/4. Therefore, the reference angle is  =  – (3/4) = /4, as shown in Figure 4.42. Because the cosecant is positive in Quadrant II, you have cont’d Figure 4.42