y
180◦, and 270◦.
90◦
(0, 1)
Solution: These angles are different from the angles we have con-
sidered so far, in that the terminal sides lie along either the x-axis
or the y-axis. So unlike the previous examples, we do not have any
180◦
0◦
right triangles to draw. However, the values of the trigonometric
x
(
0
functions are easy to calculate by picking the simplest points on their
−1,0)
(1, 0)
terminal sides and then using the definitions in formulas (1.2) and
270◦
(0, −1)
For instance, for the angle 0◦ use the point (1, 0) on its terminal
side (the positive x-axis), as in Figure 1.4.6. You could think of the
Figure 1.4.6
line segment from the origin to the point (1, 0) as sort of a degenerate
right triangle whose height is 0 and whose hypotenuse and base have
the same length 1. Regardless, in the formulas we would use r = 1, x = 1, and y = 0. Hence:
y
0
x
1
y
0
sin 0◦ =
=
= 0
cos 0◦ =
=
= 1
tan 0◦ =
=
= 0
r
1
r
1
x
1
r
1
r
1
x
1
csc 0◦ =
=
= undefined
sec 0◦ =
=
= 1
cot 0◦ =
=
= undefined
y
0
x
1
y
0
Note that csc 0◦ and cot 0◦ are undefined, since division by 0 is not allowed.
Similarly, from Figure 1.4.6 we see that for 90◦ the terminal side is the positive y-axis, so use the
point (0, 1). Again, you could think of the line segment from the origin to (0, 1) as a degenerate right
triangle whose base has length 0 and whose height equals the length of the hypotenuse. We have
r = 1, x = 0, and y = 1, and hence:
y
1
x
0
y
1
sin 90◦ =
=
= 1
cos 90◦ =
=
= 0
tan 90◦ =
=
= undefined
r
1
r
1
x
0
r
1
r
1
x
0
csc 90◦ =
=
= 1
sec 90◦ =
=
= undefined
cot 90◦ =
=
= 0
y
1
x
0
y
1
Likewise, for 180◦ use the point (−1,0) so that r = 1, x = −1, and y = 0. Hence:
y
0
x
−1
y
0
sin 180◦ =
=
= 0
cos 180◦ =
=
= −1
tan 180◦ =
=
= 0
r
1
r
1
x
−1
r
1
r
1
x
−1
csc 180◦ =
=
= undefined
sec 180◦ =
=
= −1
cot 180◦ =
=
= undefined
y
0
x
−1
y
0
Lastly, for 270◦ use the point (0, −1) so that r = 1, x = 0, and y = −1. Hence:
y
−1
x
0
y
−1
sin 270◦ =
=
= −1
cos 270◦ =
=
= 0
tan 270◦ =
=
= undefined
r
1
r
1
x
0
r
1
r
1
x
0
csc 270◦ =
=
= −1
sec 270◦ =
=
= undefined
cot 270◦ =
=
= 0
y
−1
x
0
y
−1
Trigonometric Functions of Any Angle • Section 1.4
29
The following table summarizes the values of the trigonometric functions of angles be-
tween 0◦ and 360◦ which are integer multiples of 30◦ or 45◦:
Table 1.3
Table of trigonometric function values
Angle
sin
cos
tan
csc
sec
cot
0◦
0
1
0
undefined
1
undefined
30◦
1
3
1
2
2
3
2
2
3
3
45◦
1
1
1
2
2
1
2
2
60◦
3
1
3
2
2
1
2
2
3
3
90◦
1
0
undefined
1
undefined
0
120◦
3
2
− 12
− 3
2
−2
− 1
3
3
135◦
1
− 1
−1
2
− 2
−1
2
2
150◦
1
2
2
− 3
2
− 1
− 2
− 3
3
3
180◦
0
−1
0
undefined
−1
undefined
210◦
− 1
1
3
2
− 3
2
−2
− 2
3
3
225◦
− 1
− 1
1
− 2
− 2
1
2
2
240◦
− 3
3
2
− 12
− 2
−2
1
3
3
270◦
−1
0
undefined
−1
undefined
0
300◦
− 3
1
2
2
2
− 3
− 2
− 1
3
3
315◦
− 1
1
−1
− 2
2
−1
2
2
330◦
− 1
3
2
2
− 1
−2
2
− 3
3
3
Since 360◦ represents one full revolution, the trigonometric function values repeat every
360◦. For example, sin 360◦ = sin 0◦, cos 390◦ = cos 30◦, tan 540◦ = tan 180◦, sin (−45◦) =
sin 315◦, etc. In general, if two angles differ by an integer multiple of 360◦ then each trigono-
metric function will have equal values at both angles. Angles such as these, which have the
same initial and terminal sides, are called coterminal.
In Examples 1.20-1.22, we saw how the values of trigonometric functions of an angle θ
larger than 90◦ were found by using a certain acute angle as part of a right triangle. That
acute angle has a special name: if θ is a nonacute angle then we say that the reference
angle for θ is the acute angle formed by the terminal side of θ and either the positive or
30
Chapter 1 • Right Triangle Trigonometry
§1.4
negative x-axis. So in Example 1.20, we see that 60◦ is the reference angle for the nonacute
angle θ = 120◦; in Example 1.21, 45◦ is the reference angle for θ = 225◦; and in Example
1.22, 30◦ is the reference angle for θ = 330◦.
Example 1.24
Let θ
y
= 928◦.
(a) Which angle between 0◦ and 360◦ has the same terminal side
(and hence the same trigonometric function values) as θ ?
208◦
(b) What is the reference angle for θ ?
x
0
Solution: (a) Since 928◦ = 2 × 360◦ + 208◦, then θ has the same ter-
28◦
minal side as 208◦, as in Figure 1.4.7.
928◦
(b) 928◦ and 208◦ have the same terminal side in QIII, so the refer-
ence angle for θ = 928◦ is 208◦ − 180◦ = 28◦.
Figure 1.4.7
Example 1.25
Suppose that cos θ = − 4 . Find the exact values of sin θ and tan θ.
5
Solution: We can use a method similar to the one used to solve Example 1.8 in Section 1.2. That is,
draw a right triangle and interpret cos θ as the ratio adjacent of two of its sides. Since cos θ
, we
hypotenuse
= − 45
can use 4 as the length of the adjacent side and 5 as the length of the hypotenuse. By the Pythagorean
Theorem, the length of the opposite side must then be 3. Since cos θ is negative, we know from Figure
1.4.5 that θ must be in either QII or QIII. Thus, we have two possibilities, as shown in Figure 1.4.8
below:
y
y
(−4,3)
θ
4
5
x
3
θ
0
3
x
0
5
4
(−4,−3)
(a) θ in QII
(b) θ in QIII
Figure 1.4.8
cos θ = − 45
When θ is in QII, we see from Figure 1.4.8(a) that the point (−4,3) is on the terminal side of θ, and so we have x
y
y
= −4, y = 3, and r = 5. Thus, sin θ =
and tan θ
.
r = 3
5
= x = 3
−4
When θ is in QIII, we see from Figure 1.4.8(b) that the point (−4,−3) is on the terminal side of θ, and so we have x
y
y
= −4, y = −3, and r = 5. Thus, sin θ =
and tan θ
.
r = −3
5
= x = −3
−4 = 3
4
Thus, either sin θ = 3 and tan θ
or sin θ
and tan θ
.
5
= − 34
= − 35
= 34
Since reciprocals have the same sign, csc θ and sin θ have the same sign, sec θ and cos θ
have the same sign, and cot θ and tan θ have the same sign. So it suffices to remember the
signs of sin θ, cos θ, and tan θ:
Trigonometric Functions of Any Angle • Section 1.4
31
For an angle θ in standard position and a point ( x, y) on its terminal side:
(a) sin θ has the same sign as y
(b) cos θ has the same sign as x
(c) tan θ is positive when x and y have the same sign
(d) tan θ is negative when x and y have opposite signs
Exercises
For Exercises 1-10, state in which quadrant or on which axis the given angle lies.
1. 127◦
2. −127◦
3. 313◦
4. −313◦
5. −90◦
6. 621◦
7. 230◦
8. 2009◦
9. 1079◦
10. −514◦
11. In which quadrant(s) do sine and cosine have the same sign?
12. In which quadrant(s) do sine and cosine have the opposite sign?
13. In which quadrant(s) do sine and tangent have the same sign?
14. In which quadrant(s) do sine and tangent have the opposite sign?
15. In which quadrant(s) do cosine and tangent have the same sign?
16. In which quadrant(s) do cosine and tangent have the opposite sign?
For Exercises 17-21, find the reference angle for the given angle.
17. 317◦
18. 63◦
19. −126◦
20. 696◦
21. 275◦
For Exercises 22-26, find the exact values of sin θ and tan θ when cos θ has the indicated value.
22. cos θ = 1
23. cos θ
24. cos θ
26. cos θ
2
= − 12
= 0
25. cos θ = 25
= 1
For Exercises 27-31, find the exact values of cos θ and tan θ when sin θ has the indicated value.
27. sin θ = 1
28. sin θ
29. sin θ
31. sin θ
2
= − 12
= 0
30. sin θ = − 23
= 1
For Exercises 32-36, find the exact values of sin θ and cos θ when tan θ has the indicated value.
32. tan θ = 1
33. tan θ
34. tan θ
36. tan θ
2
= − 12
= 0
35. tan θ = 5
12
= 1
For Exercises 37-40, use Table 1.3 to answer the following questions.
37. Does sin 180◦ + sin 45◦ = sin 225◦ ?
38. Does tan 300◦ − tan 30◦ = tan 270◦ ?
39. Does cos 180◦ − cos 60◦ = cos 120◦ ?
40. Does cos 240◦ = (cos 120◦)2 − (sin 120◦)2 ?
41. Expand Table 1.3 to include all integer multiples of 15◦. See Example 1.10 in Section 1.2.
32
Chapter 1 • Right Triangle Trigonometry
§1.5
1.5 Rotations and Reflections of Angles
Now that we know how to deal with angles of any measure, we will take a look at how certain
geometric operations can help simplify the use of trigonometric functions of any angle, and
how some basic relations between those functions can be made. The two operations on which
we will concentrate in this section are rotation and reflection.
To rotate an angle means to rotate its terminal side around the origin when the angle
is in standard position. For example, suppose we rotate an angle θ around the origin by 90◦
in the counterclockwise direction. In Figure 1.5.1 we see an angle θ in QI which is rotated
by 90◦, resulting in the angle θ + 90◦ in QII. Notice that the complement of θ in the right
triangle in QI is the same as the supplement of the angle θ + 90◦ in QII, since the sum of θ,
its complement, and 90◦ equals 180◦. This forces the other angle of the right triangle in QII
to be θ.
y
( x, y)
(− y, x)
r
r
θ + 90◦
θ
y
x
90◦
θ
x
y
x
Figure 1.5.1
Rotation of an angle θ by 90◦
Thus, the right triangle in QI is similar to the right triangle in QII, since the triangles
have the same angles. The rotation of θ by 90◦ does not change the length r of its terminal
side, so the hypotenuses of the similar right triangles are equal, and hence by similarity
the remaining corresponding sides are also equal. Using Figure 1.5.1 to match up those
corresponding sides shows that the point (− y, x) is on the terminal side of θ +90◦ when ( x, y)
is on the terminal side of θ. Hence, by definition,
x
− y
x
sin ( θ + 90◦) =
= cos θ , cos ( θ + 90◦) =
= −sin θ , tan ( θ + 90◦) =
= −cot θ .
r
r
− y
Though we showed this for θ in QI, it is easy (see Exercise 4) to use similar arguments for
the other quadrants. In general, the following relations hold for all angles θ:
sin ( θ + 90◦) = cos θ
(1.4)
cos ( θ + 90◦) = −sin θ
(1.5)
tan ( θ + 90◦) = −cot θ
(1.6)
Rotations and Reflections of Angles • Section 1.5
33
Example 1.26
Recall that any nonvertical line in the x