Download Evaluation of Trig Functions

3. EVALUATION OF TRIGONOMETRIC
FUNCTIONS
In this section, we obtain values of the trigonometric functions for quadrantal angles, we
introduce the idea of reference angles, and we discuss the use of a calculator to evaluate
trigonometric functions of general angles.
In Definition 2.1, the domain of each trigonometric function consists of all angles θ for
which the denominator in the corresponding ratio is not zero. Because r > 0, it follows
that sin θ = y/r and cos θ = x/r are defined for all angles θ . However, tan θ = y/x and
sec θ = r/x are not defined when the terminal side of θ lies anlong the y axis (so that x =
0). Likewise, cot θ = x/y and csc θ = r/y are not defined when the terminal side of θ lies
along the x axis (so that y = 0). Therefore, when you deal with a trigonometric function of
a quadrantal angle, you must check to be sure that the function is actually defined for that
angle.
Example 3.1 ---------------------------- -----------------------------------------------------------Find the values (if they are defined) of the six trigonometric functions for the quadrantal
angle θ = 90° (or θ = π 2 ).
In order to use Definition 1, we begin by choosing any point ( 0 , y ) with y > 0, on the terminal side of the
90° angle (Figure 1). Because x = 0, it follows that tan 90° and sec 90° are undefined. Since y > 0, we
have
r=
x2 + y2 =
y
y
=
=1
r
y
r
y
csc 90° =
=
=1
y
y
Therefore, sin 90° =
02 + y 2 =
y 2 = y = y.
0
x
=
=0
r
y
x
0
= 0.
cot 90° =
=
y
y
cos 90° =
_______________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________
The values of the trigonometric functions for other quadrantal angles are found in
a similar manner. The results appear in Table 3.1. Dashes in the table indicate that the
function is undefined for that angle.
Table 3.1
θ degrees
θ radians
sin θ
cos θ
tan θ
cot θ
sec θ
csc θ
0°
0
0
1
0
––
1
––
1
0
––
0
––
1
90°
π
2
180°
π
0
–1
0
––
–1
––
270°
3π
2
–1
0
––
0
––
–1
360°
2π
0
1
0
––
1
––
18
It follows from Definition 2.1 that the values of each of the six trigonometric functions
remain unchanged if the angle is replaced by a coterminal angle. If an angle exceeds one
revolution or is negative, you can change it to a nonnegative coterminal angle that is less
than one revolution by adding or subtracting an integer multiple of 360° (or 2π radians).
For instance,
sin 450° = sin( 450° − 360° ) = sin 90° = 1.
sec 7π = sec ( 7π − ( 3 × 2π ) ) = sec π = –1.
cos (− 660°) = cos (− 660° + (2 × 360°) ) = cos 60° =
1
2
.
In Examples 3.2 and 3.3, replace each angle by a nonnegative coterminal angle that is
less than on revolution and then find the values of the six trigonometric functions (if they
are defined).
Example 3.2 ---------------------------- -----------------------------------------------------------θ = 1110°
By dividing 1110 by 360, we find that the largest integer multiple of 360° that is less than 1110° is
3 × 360° = 1080° . Thus,
1110° – ( 3 × 360° ) = 1110° – 1080° = 30° .
(Or we could have started with 1110° and repeatedly subtracted 360° until we obtained 30° .) It follows
that
1
sin 1110° = sin 30° =
csc 1110° = csc 30° = 2
2
3
2 3
sec 1110° = sec 30° =
cos 1110° = cos 30° =
2
3
3
cot 1110° = cot 30° = 3
tan 1110° = tan 30° =
3
___________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________
Example 3.3 ---------------------------- -----------------------------------------------------------θ =–
5π
2
We repeatedly add 2 π to –
–
–
5π
until we obtain a nonnegative coterminal angle:
2
5π
π
+ 2π = –
2
2
π
2
+ 2π =
(still negative)
3π
.
2
Therefore, by Table 3.1 for quadrantal angles,
⎛ 5π ⎞
⎛ 3π ⎞
⎛ 5π ⎞
⎛ 3π ⎞
cot ⎜ − ⎟ = cot ⎜ ⎟ = 0
sin ⎜ −
⎟ = sin ⎜ ⎟ = –1
2
2
2
⎝
⎠
⎝ ⎠
⎝
⎠
⎝ 2 ⎠
⎛ 5π ⎞
⎛ 3π ⎞
⎛ 5π ⎞
⎛ 3π ⎞
csc ⎜ −
cos ⎜ −
⎟ = cos ⎜ ⎟ = 0
⎟ = csc ⎜ ⎟ = –1
⎝ 2 ⎠
⎝ 2 ⎠
⎝ 2 ⎠
⎝ 2 ⎠
⎛ 5π ⎞
⎛ 5π ⎞
and both tan ⎜ −
⎟ and sec ⎜ −
⎟ are undefined.
⎝ 2 ⎠
⎝ 2 ⎠
______________________________________________________________________________________
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Table 3.1
θ degrees
30°
θ radians
π
6
π
45°
4
π
60°
sin θ
1
2
cos θ
tan θ
cot θ
sec θ
csc θ
3
2
3
3
3
2 3
3
2
2
2
2
2
1
2
1
1
2
2
3
3
3
2
2 3
3
3
2
3
Figure 3.2
y
y
θ
θR
θ = θR
O
O
x
(a)
y
x
(b)
y
θ R = θ ‐ 180°
θ R = θ ‐ π
θ
θR
θ R = 180° ‐ θ
θ R = π ‐ θ
θ R = 360° ‐ θ
θ R = 2π ‐ θ
θ
O
O
x
x
θR
(c)
(d)
20
Example 3.4 ---------------------------- -----------------------------------------------------------Find the reference angle θ R for each angle θ .
3π
(a) θ = 60°
(b) θ =
(c) θ = 210°
4
(a) By Figure 3.2(a), θ R = θ = 60° .
(d) θ =
5π
.
3
3π
π
=
.
4
4
(c) By Figure 3.2(c), θ R = θ – 180° = 210° – 180° = 30° .
π
5π
(d) By Figure 3.2(c), θ R = 2 π – θ = 2 π –
=
.
3
3
______________________________________________________________________________________
(b) By Figure 3.2(b), θ R = π – θ = π –
The value of any trigonometric function of any angle θ is the same as the value of the
function for the reference angle, θ R , except possibly for a change of algebraic sign.
Thus,
sin θ = ± sin θ R ,
cos θ = ± cos θ R ,
and so forth. You can always determine the correct algebraic sign by considering the quadrant in which
θ lies.
Section 3 Problems---------------------- -----------------------------------------------------------In problems 1 and 2, find the values (if they are defined) of the six trigonometric
functions of the given quadrantal angles. (Do not use a calculator.)
1. (a) 0°
(b) 180°
(c) 270°
(d) 360° .
[When you have finished, compare your answers with the results in Table 3.1]
2. (a) 5 π
(b) 6 π
(c) –7 π
(d)
5π
2
(e)
7π
.
2
In Problems 3 to 14, replace each angle by a nonnegative coterminal angle that is less
than one revolution and then find the exact values of the six trigonometric functions (if
they are defined) for the angle.
4. 810°
3. 1440°
5. 900°
6. – 220°
7. 750°
8. 1845°
9. – 675°
11. 5 π
13.
17π
3
19π
2
25π
12.
6
31π
14. –
4
10.
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15. What happens when you try to evaluate tan 900° on a calculator? [Try it.]
16. Let θ be a quadrant III angle in standard position and let θ R be its reference angle.
Show that the value of any trigonometric function of θ is the same as the value of
θ R , except possibly for a change of algebraic sign. Repeat for θ in quadrant IV.
In problems 17 to 36, find the reference angle θ R for each angle θ , and then find the
exact values of the six trigonometric functions of θ .
17. θ = 150°
18. θ = 120°
19. θ = 240°
20. θ = 225°
22. θ = 675°
21. θ = 315°
5π
6
13π
=–
6
53π
=
6
9π
=
4
50π
=–
3
147π
=–
4
= – 5370°
23. θ = – 150°
24. θ = –
25. θ = – 60°
26. θ
27. θ = –
29. θ
31. θ
33. θ
35. θ
π
4
2π
=–
3
7π
=
4
11π
=
3
= – 420°
28. θ
30. θ
32. θ
34. θ
36. θ
37. Complete the following tables. (Do not use a calculator.)
θ degrees
210°
225°
240°
300°
315°
330°
θ radians
sin θ
cos θ
7π
6
5π
4
4π
3
5π
3
7π
4
11π
6
22
tan θ
θ degrees
210°
225°
240°
300°
315°
330°
θ radians
cot θ
sec θ
csc θ
7π
6
5π
4
4π
3
5π
3
7π
4
11π
6
38. A calculator is set in radian mode. π is entered and the sine (SIN) key is pressed.
The display shows – 4.1 × 10 -10 . But we know that sin π = 0. Explain.
In problems 59 to 62, use a calculator to verify that the equation is true for the indicated
value of the angle θ .
59. tan θ =
sin θ
cos θ
for θ = 35° .
60. (cos θ )(tan θ ) = sin θ
for θ =
61. cos 2 θ + sin 2 θ = 1
for θ =
5π
7
5π
3
62. 1 + tan 2 θ = sec 2 θ for θ = 17.75°
63. Verify that for θ = 0° , 30° , 45° , 60° , 90° , we have sin θ =
2
4
0
1
3
, ,
,
, and
2
2
2
2
2
respectively. [Although there is no theoretical significance to this pattern, people often
use it as a memory aid to help recall these values of sin θ .]
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