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2. TRIGONOMETRIC FUNCTIONS OF
GENERAL ANGLES
In order to extend the definitions of the six trigonometric functions to general angles, we
shall make use of the following ideas: In a Cartesian coordinate system, an angle α is
said to be in standard position if its vertex is at the origin O and its initial side
coincides with the positive x axis (Figure 2.1). An angle is said to be in a certain
quadrant if, when the angle is in standard position, the terminal side lies in that quadrant.
For instance, a 65° angle lies in quadrant I or simply that it is a quadrant I angle. As
Figure 2b shows, an angle of ‐187° is a quadrant II angle.
If the terminal side of an angle in standard position lies along either the x axis or the y
axis, then the angle is called quadrantal. For example, ‐360° , ‐270° , ‐180° , ‐90° , 0° ,
90° , 180° , 270° , 360° are quadrantal angles. Evidently, an angle is quadrantal if and only
if its measure is an integer multiple of 90° ( or
Figure 2.1
π
2
radians).
α is in standard position
y
terminal
side
α
x
initial side
Figure 2.2
(a)
(b) quadrant II angle
quadrant I angle
y
y
terminal
side
terminal
side
65°
x
x
‐ 187°
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Definition 2.1: Trigonometric Functions of a General Angle
Let θ be an angle in standard position and suppose that ( x , y ) is any point other than
( 0 , 0 ) on the terminal side of θ (Figure 2.3). If r = x 2 + y 2 is the distance between
( x , y ) and ( 0 , 0 ), then the six trigonometric functions of θ are defined by
Figure 2.3
y
r
x
cos θ =
r
y
tan θ =
x
r
y
r
sec θ =
x
x
cot θ =
y
sin θ =
csc θ =
y
.
(x,y)
θ
r
provided that the denominators are not zero.
O
Using similar triangles, you can see that the values of the six trigonometric functions in
Definition 2.1 depend only on the angle θ and not on the choice of the point (x, y) on the
terminal side of θ .
Example 2.1 ---------------------------- -----------------------------------------------------------Evaluate the six trigonometric functions of the angle θ in standard position if the
terminal side of θ contains the point (x, y) = ( 2, -1).
Here, x = 2, y = -1, and
r=
x2 + y2 =
2 2 + (−1) 2 =
5.
Thus,
sin θ =
y
r
=
x
r
=
−1
=
5
− 5
5
2 5
5
5
−1
1
y
tan θ =
=
= −
x
2
2
cos θ =
2
=
csc θ =
5
r
=
=
y
−1
5
5
r
=
x
2
2
x
cot θ =
=
= −2 .
y
−1
sec θ =
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You can determine the algebraic signs of the trigonometric functions for angles in the
various quadrants by recalling the algebraic signs of x and y in these quadrants and
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x
y
is
r
positive in quadrants I and II (where both y and r are positive), and it is negative in
quadrants III and IV (where y is negative and r is positive). By proceeding in a similar
way, you can determine the signs of the remaining trigonometric functions in the various
quadrants and thus confirm the results in Table 2.1.
remembering that r is always positive. For instance, as Figure 2.4 shows, sin θ =
Figure 2.4
y
y>0
sin θ =
y>0
y
sin θ =
>0
>0
r
Quadrant I
r
Quadrant II
O
y<0
sin θ =
y
y<0
y
sin θ =
<0
x
y
<0
r
Quadrant IV
r
Quadrant III
Table 2.1
Quadrant
Containing θ
I
II
III
IV
Positive
Functions
All
sin θ , csc θ
tan θ , cot θ
cos θ , sec θ
Negative
Functions
None
cos θ , sec θ , tan θ , cot θ
sin θ , csc θ , cos θ , sec θ
sin θ , csc θ , tan θ , cot θ
Example 2.2 ---------------------------- -----------------------------------------------------------Find the quadrant in which θ lies if tan θ > 0 and sin θ < 0.
This example can be worked by using Table 2.1; however, rather than relying on the table, we prefer to
reason as follows: Let ( x , y ) be a point other than the origin on the terminal side of θ (in standard
y
position). Because tan θ =
> 0, we see that x and y have the same algebraic sign. Furthermore, since
x
y
sin θ =
< 0, it follows that y < 0. Because x < 0 and y < 0, the angle is in quadrant III.
r
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Reciprocal Identities
If θ is an angle for which the functions are defined, then:
(i) csc θ =
1
sin θ
(ii) sec θ =
1
cos θ
(iii) cot θ =
1
tan θ
.
Quotient Identities
If θ is an angle for which the functions are defined, then:
tan θ =
sin θ
cos θ
and
cos θ
sin θ
cot θ =
.
Example 2.3 ---------------------------- -----------------------------------------------------------If sin θ =
1
3
and cos θ =
−2 2
3
, find the values of the other four trigonometric functions
of θ .
tan θ =
sec θ =
cot θ =
1
sin θ
3
=
cos θ
−2 2
= −
2 2
3
1
1
=
cos θ
−2 2
1
1
=
tan θ
− 2
1
= −
3
= −
2 2
3
= −
4
1
1
=
= 3.
csc θ =
1
sin θ
3
= −
4
2
4
3 2
4
= −2 2
2
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By using the reciprocal and quotient identities, you can quickly recall the
algebraic signs of the secant, cosecant, tangent, and cotangent in the four quadrants
(Table 1), if you know the algebraic signs of the sine and cosine in these quadrants.
Another important identity is derived as follows: Again suppose that θ is an angle
in standard position and that ( x , y ) is a point other than the origin on the terminal side
2
2
2
of θ (Figure 9). Because r = x 2 + y 2 , we have x + y = r , so
(cos θ ) 2 + (sin θ ) 2
2
2
x2
y2
r2
x
y
= ⎛⎜ ⎞⎟ + ⎛⎜ ⎞⎟ = 2 + 2 = 2 = 1.
⎝r⎠
⎝r⎠
r
r
r
(cos θ ) 2 + (sin θ ) 2 = 1
The relationship:
is called the fundamental Pythagorean identity because its derivation involves the fact
2
2
2
that x + y = r , which is a consequence of the Pythagorean theorem.
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The fundamental Pythagorean identity is used quite often, and it would be
2
2
bothersome to write the parentheses each time for (cosθ ) and (sinθ ) ; yet, if the
parentheses were simply omitted, the resulting expressions would be misunderstood. (For
2
instance, cos θ is usually understood to mean the cosine of the square of θ .) Therefore,
2
2
2
2
it is customary to write cos θ and sin θ to mean (cosθ ) and (sinθ ) .
Similar notation is used for the remaining trigonometric functions and for powers other
4
4
n
n
than 2. Thus, cot θ means ( cot θ ) , sec θ means ( sec θ ) , and so forth. With this
notation, the fundamental Pythagorean identity becomes
cos 2 θ
+ sin 2 θ = 1.
Actually, there are three Pythagorean identities – the fundamental identity and
two others derived from it.
Pythagorean Identities
If θ is an angle for which the functions are defined, then:
(i) cos 2 θ + sin 2 θ = 1
(ii) 1+ tan 2 θ = sec 2 θ
(iii) 1+ cot 2 θ = csc 2 θ
We already proved (i). To prove (ii), we divide both sides of (i) by cos 2 θ to obtain
1+
sin 2 θ
cos 2 θ
=
1
cos 2 θ
2
2
sin θ ⎞
⎛ 1 ⎞
1 + ⎛⎜
⎟ ,
⎟ = ⎜
or
⎝ cos θ ⎠
⎝ cos θ ⎠
provided that cos θ ≠ 0 . Since
sin θ
= tan θ
cos θ
1
cos θ
and
= sec θ ,
we have that
1 + tan 2 θ = sec 2 θ .
2
Identity (iii) is proved by dividing both sides of (i) by sin θ .
Example 2.4 ---------------------------- -----------------------------------------------------------The value of one of the trigonometric functions of an angle θ is given along with the
information about the quadrant in which θ lies, Find the values of the other five
trigonometric functions of θ :
13
( a ) sin θ =
5
, θ
13
in quadrant II.
2
2
By the fundamental Pythagorean identity, cos θ + sin θ = 1, so
2
⎛5⎞
cos 2 θ = 1 − sin θ = 1 − ⎜ ⎟
⎝ 13 ⎠
2
= 1−
25
169 − 25
144
=
=
.
169
169
169
144
12
= ±
.
169
13
Because θ is in quadrant II, we know that cos θ is negative; hence,
12
.
cos θ = 13
It follows that
5
sin θ
1
12
1
13 = − 5
tan θ =
=
cot θ =
=
= −
-5
cos θ
12
tan θ
5
− 12
12
13
1
1
13
1
13
1
=
= −
csc θ =
=
=
sec θ =
5
− 12
cos θ
5
12
sin θ
13
13
Therefore, cos θ = ±
( b ) tan θ = ‐ 8
and sin θ < 0.
7
Because tan θ < 0 only in quadrants II and IV, and sin θ < 0 only in quadrants III and IV, it follows that
θ must be in quadrant IV. By part (ii) sec 2 θ = 1 + tan 2 θ , so
( )2
sec θ = ± 1+ tan 2 θ = ± 1 + - 87
= ± 1+
64
49
= ±
Since θ is in quadrant IV, sec θ > 0; hence,
sec θ =
113
.
7
Because sec θ =
1
,
cos θ
it follows that
1
cos θ =
=
secθ
1
Now, tan θ =
so
Finally,
113
113
= ±
.
49
7
113
sin θ
cos θ
=
7
=
113
7
⎛ 8⎞
sin θ = (tan θ )(cos θ ) = ⎜ − ⎟
⎝ 7⎠
csc θ =
and cot θ =
1
=
sin θ
1
=
tan θ
7 113
.
113
1
‐ 8
7
1
- 8 113
= −
⎛ 7 113 ⎞
⎜
⎟ = − 8 113 .
⎜ 113 ⎟
113
⎝
⎠
= −
113
113
8
7
.
8
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In the applications of trigonometry, and especially in calculus, it is often necessary to
make trigonometric calculations, as we have done in this section, without the use of
calculators or tables.
Section 2 Problems--------------- ------- -----------------------------------------------------------In problems 1 to 10, sketch two coterminal angles α and β in standard position whose
terminal side contains the given point. Arrange it so that α is positive, β is negative, and
neither angle exceeds one revolution. In each case, name the quadrant in which the angle
lies, or indicate that the angle is quadrantal.
4
3
−4
1. ( 1 , 2 )
2. ( 2 , − )
3. ( −5 , 0 )
5. ( 5 , −3 )
4. ( −3 ,
6. ( 0 , 4 )
7. ( −1 , 1 )
8. (
9. ( −
3
4
, −
3
4
)
2
2
)
,0)
10. ( 0 , −3 )
In problems 11 to 18, specify and sketch three angles that are coterminal with the given
angle in standard position.
11. 60°
12. −15°
13. −
14.
π
4
15. − 612°
17.
5π
6
4π
3
7π
16. −
4
18. 1440 °
In problems 16 to 28, evaluate the six trigonometric functions of the angle θ in standard
position if the terminal side of θ contains the given point ( x , y ). [Do not use a calculator
– leave all answers in the form of a fraction or an integer.] In each case, sketch one of the
coterminal angles θ .
19. ( 4 , 3 )
20. ( 2 , 7 )
21. (–5, 12 )
22. (–2, 4 )
23. (–3, –4 )
24. ( 1 , 1 )
25. ( 7 , 3 )
26. ( 1 , –3 )
27. (– 3 , – 2 )
28. ( 20 , 21 )
29. Is there any angle θ for which sin θ =
5
4
? Explain.
15
30. Using similar triangles, show that the values of the six trigonometric functions in
Definition 2.1 depend only on the angle θ and not on the choice of the point
( x , y ) on the terminal side of θ .
31. In each case, assume that θ is an angle in standard position and find the quadrant in
which it lies.
(a) tan θ > 0 and
sec θ > 0
(b) sin θ > 0 and
sec θ < 0
(c) sin θ > 0 and
cos θ < 0
tan θ < 0
(d) sec θ > 0 and
(e) tan θ > 0 and
csc θ < 0
csc θ < 0
(f) cos θ < 0 and
(g) sec θ > 0 and
cot θ < 0
sin θ > 0
(h) cot θ > 0 and
32. Is there any angle θ for which sin θ > 0 and csc θ < 0? Explain.
33. Give the algebraic sign of each of the following.
(a) cos 163°
(b) sin 211°
(c) sec (− 355 o )
28π ⎞
⎟
⎝ 45 ⎠
38π ⎞
sec ⎛⎜
⎟
⎝ 15 ⎠
(e) cot ⎛⎜ −
(g)
7π ⎞
⎟
⎝ 4 ⎠
5π
csc ⎛⎜ − ⎞⎟
⎝ 12 ⎠
(d) tan ⎛⎜
(f)
34. If θ is an angle for which the functions are defined, show that
sec θ – ( sin θ )( tan θ ) = cos θ .
35. If sin θ = –
5
13
(a) sec θ
and cos θ =
(b) csc θ
12
13
, use the reciprocal and quotient identities to find
(c) tan θ
16
(d) cot θ .
36. If sec θ = –
5
4
(a) sin θ
5
3
and csc θ = – , use the reciprocal and quotient identities to find
(b) cos θ
(c) tan θ
(d) cot θ .
In Problems 37 to 48, the values of one of the trigonometric functions of an angle θ is
given along with information about the quadrant (Q) in which θ lies. Find the values of
the other five trigonometric functions of θ .
4
,
3
12
cos θ = ,
13
3
sin θ = – ,
4
3
sin θ = ,
4
4
cos θ = ,
7
1
cos θ = ,
3
3
csc θ = ,
2
5
sec θ = – ,
3
4
tan θ = ,
3
37. sin θ =
θ in QI
38.
θ in QIV
39.
40.
41.
42.
43.
44.
45.
46. tan θ = 2,
47.
48.
12
cot θ = – ,
5
25
csc θ = – ,
7
θ in QIII
θ not in QI
sin θ < 0
θ not in QI
θ in QI
θ in QIII
θ in QI
sin θ < 0
csc θ > 0
sec θ < 0
17