Section 5.2 Trigonometric Functions of Real Numbers
The Trigonometric Functions
EXAMPLE: Use the Table below to find the six trigonometric functions of each given real
number t.
π
π
(a) t =
(b) t =
3
2
1
EXAMPLE: Use the Table below to find the six trigonometric functions of each given real
number t.
π
π
(a) t =
(b) t =
3
2
Solution:
(a) From the Table,√we see that the terminal point determined by
t =√
π/3 is P (1/2, 3/2). Since the coordinates are x = 1/2 and
y = 3/2, we have
√
√
π
3
3/2 √
π
1
π
sin =
cos =
tan =
= 3
3
2
3
2
3
1/2
√
√
π
3
2 3
π
π
1/2
csc =
=
sec = 2
cot = √
3
3
3
3
3
3/2
(b) The terminal point determined by π/2 is P (0, 1). So
π
π
1
π
0
π
cos = 0
csc = = 1
cot = = 0
sin = 1
2
2
2
1
2
1
But tan π/2 and sec π/2 are undefined because x = 0 appears in the denominator in each of
their definitions.
π
.
4
Solution:
the terminal point
determined by t = π/4 is
√
√From the Table above, we see that √
√
P ( 2/2, 2/2). Since the coordinates are x = 2/2 and y = 2/2, we have
√
√
√
π
2
2
2/2
π
π
sin =
=1
cos =
tan = √
4
2
4
2
4
2/2
√
π √
π
π √
2/2
csc = 2
sec = 2
cot = √
=1
4
4
4
2/2
EXAMPLE: Find the six trigonometric functions of each given real number t =
2
Values of the Trigonometric Functions
EXAMPLE:
π
π
(a) cos > 0, because the terminal point of t = is in Quadrant I.
3
3
(b) tan 4 > 0, because the terminal point of t = 4 is in Quadrant III.
(c) If cos t < 0 and sin t > 0, then the terminal point of t must be in Quadrant II.
EXAMPLE: Determine the sign of each function.
7π
(b) tan 1
(a) cos
4
Solution:
(a) Positive
(b) Positive
EXAMPLE: Find each value.
π
2π
(a) cos
(b) tan −
3
3
(c) sin
3
19π
4
EXAMPLE: Find each value.
π
19π
2π
(b) tan −
(c) sin
(a) cos
3
3
4
Solution:
(a) Since
2π
3π − π
3π π
π
=
=
− =π−
3
3
3
3
3
the reference number for 2π/3 is π/3 (see Figure (a) below) and the terminal point of 2π/3 is
in Quadrant II. Thus cos(2π/3) is negative and
(b) The reference number for −π/3 is π/3 (see Figure (b) below). Since the terminal point of
−π/3 is in Quadrant IV, tan(−π/3) is negative. Thus
(c) Since
19π
20π − π
20π π
π
=
=
− = 5π −
4
4
4
4
4
the reference number for 19π/4 is π/4 (see Figure (c) below) and the terminal point of 19π/4
is in Quadrant II. Thus sin(19π/4) is positive and
EXAMPLE: Find each value. 2π
4π
(a) sin
(b) tan −
3
3
(c) cos
4
14π
3
EXAMPLE: Find each value. 4π
2π
(b) tan −
(a) sin
3
3
(c) cos
14π
3
Solution:
(a) Since
2π
3π − π
3π π
π
=
=
− =π−
3
3
3
3
3
the reference number for 2π/3 is π/3 and the terminal point of 2π/3 is in Quadrant II. Thus
sin(2π/3) is positive and
√
2π
3
π
sin
= sin =
3
3
2
(b) Since
4π
3π + π
3π π
π
=−
=−
− = −π −
3
3
3
3
3
the reference number for −4π/3 is π/3 and the terminal point of −4π/3 is in Quadrant II. Thus
tan (−4π/3) is negative and
π √
4π
= − tan
=− 3
tan −
3
3
−
(c) Since
14π
15π − π
15π π
π
=
=
− = 5π −
3
3
3
3
3
the reference number for 14π/3 is π/3 and the terminal point of 14π/3 is in Quadrant II. Thus
cos(14π/4) is negative and
14π
π
1
cos
= − cos = −
3
3
2
EXAMPLE: Evaluate
7π
π
(b) cos
(a) sin
3
6
(c) tan
11π
4
(d) sec
5
17π
3
(e) csc
17π
2
(f) cot
121π
6
EXAMPLE: Evaluate
7π
11π
17π
17π
121π
π
(b) cos
(c) tan
(d) sec
(e) csc
(f) cot
(a) sin
3
6
4
3
2
6
Solution:
(a) The reference number for π/3 is π/3. Since the terminal point of π/3 is in Quadrant I,
sin(π/3) is positive. Thus
√
π
3
π
sin = sin =
3
3
2
(b) Since 7π
= 6π+π
= 6π
+ π6 = π + π6 , the reference number for 7π/6 is π/6 and the terminal
6
6
6
point of 7π/6 is in Quadrant III. Thus cos(7π/6) is negative and
√
π
7π
3
= − cos = −
cos
6
6
2
(c) Since 11π
= 12π−π
= 12π
− π4 = 3π − π4 , the reference number for 11π/4 is π/4 and the
4
4
4
terminal point of 11π/6 is in Quadrant II. Thus tan(11π/4) is negative and
tan
π
11π
= − tan = −1
4
4
= 18π−π
= 18π
− π3 = 6π − π3 , the reference number for 17π/3 is π/3 and the
(d) Since 17π
3
3
3
terminal point of 17π/3 is in Quadrant IV. Thus sec(17π/3) is positive and
sec
17π
π
= sec = 2
3
3
= 16π+π
= 16π
+ π2 = 8π + π2 , the reference number for 17π/2 is π/2 and the
(e) Since 17π
2
2
2
terminal point of 17π/2 is in Quadrant I (II). Thus csc(17π/2) is positive and
csc
π
17π
= csc = 1
2
2
(f) Since 121π
= 120π+π
= 120π
+ π6 = 20π + π6 , the reference number for 121π/6 is π/6 and the
6
6
6
terminal point of 121π/6 is in Quadrant I. Thus cot(121π/6) is positive and
cot
π √
121π
= cot = 3
6
6
EXAMPLE: Use the even-odd properties of the trigonometric functions to determine each value.
π
π
π
(b) cos −
(c) csc −
(a) sin −
6
4
3
6
EXAMPLE: Use the even-odd properties of the trigonometric functions to determine each value.
π
π
π
(a) sin −
(b) cos −
(c) csc −
6
4
3
Solution: We have
√
π
π
π
1
π
2
= − sin = −
(b) cos −
= cos =
(a) sin −
6
6
2
4
4
2
√
π
2 3
π
(c) csc −
= − csc = −
3
3
3
Fundamental Identities
Proof: The reciprocal identities follow immediately from the definition. We now prove the
Pythagorean identities. By definition, cos t = x and sin t = y, where x and y are the coordinates
of a point P (x, y) on the unit circle. Since P (x, y) is on the unit circle, we have x2 + y 2 = 1.
Thus
sin2 t + cos2 t = 1
Dividing both sides by cos2 t (provided cos t 6= 0), we get
sin2 t cos2 t
1
+
=
2
2
cos t cos t
cos2 t
2 2
2
sin t
1
+1=
cos2 t
cos2 t
tan2 t + 1 = sec2 t
We have used the reciprocal identities sin t/ cos t = tan t and 1/ cos t = sec t. Similarly, dividing
both sides of the first Pythagorean identity by sin2 t (provided sin t 6= 0) gives us 1 + cot2 t =
csc2 t.
3
and t is in Quadrant IV, find the values of all the trigonometric
EXAMPLE: If cos t =
5
functions at t.
7
3
and t is in Quadrant IV, find the values of all the trigonometric
EXAMPLE: If cos t =
5
functions at t.
Solution: From the Pythagorean identities we have
sin2 t + cos2 t = 1
2
3
2
sin t +
=1
5
sin2 t = 1 −
sin t = ±
9
16
=
25
25
4
5
4
Since this point is in Quadrant IV, sin t is negative, so sin t = − . Now that we know both
5
sin t and cos t, we can find the values of the other trigonometric functions using the reciprocal
identities:
sin t = −
csc t =
4
5
1
5
=−
sin t
4
cos t =
3
5
tan t =
−4
4
sin t
= 35 = −
cos t
3
5
sec t =
1
5
=
cos t
3
cot t =
1
3
=−
tan t
4
5
and t is in Quadrant II, find the values of all the trigonometric
EXAMPLE: If cos t = −
13
functions at t.
Solution: From the Pythagorean identities we have
sin2 t + cos2 t = 1
2
5
sin t + −
=1
13
2
25
144
=
169
169
12
sin t = ±
13
sin2 t = 1 −
12
Since this point is in Quadrant II, sin t is positive, so sin t =
. Now that we know both
13
sin t and cos t, we can find the values of the other trigonometric functions using the reciprocal
identities:
sin t =
12
13
cos t = −
csc t =
1
13
=
sin t
12
sec t =
5
13
12
12
sin t
= 135 = −
cos t
5
− 13
1
5
cot t =
=−
tan t
12
tan t =
1
13
=−
cos t
5
EXAMPLE: Write tan t in terms of cos t, where t is in Quadrant III.
8
EXAMPLE: Write tan t in terms of cos t, where t is in Quadrant III.
Solution: Since tan t = sin t/ cos t, we need to write sin t in terms of cos t. By the Pythagorean
identities we have
sin2 t + cos2 t = 1
sin2 t = 1 − cos2 t
√
sin t = ± 1 − cos2 t
Since sin t is negative in Quadrant III, the negative sign applies here. Thus
√
1 − cos2 t
sin t
=−
tan t =
cos t
cos t
EXAMPLE: Write tan t in terms of cos t, where t is in Quadrant I.
Solution: Since tan t = sin t/ cos t, we need to write sin t in terms of cos t. By the Pythagorean
identities we have
sin2 t + cos2 t = 1
sin2 t = 1 − cos2 t
√
sin t = ± 1 − cos2 t
Since sin t is positive in Quadrant I, the positive sign applies here. Thus
√
1 − cos2 t
sin t
=
tan t =
cos t
cos t
EXAMPLE: Write cos t in terms of tan t, where t is in Quadrant II.
Solution: Since tan t = sin t/ cos t, we need to write sin t in terms of cos t. By the Pythagorean
identities we have
sin2 t + cos2 t = 1
sin2 t = 1 − cos2 t
so tan2 t =
1 − cos2 t
sin2 t
=
. Multiplying both sides by cos2 t, we get
cos2 t
cos2 t
cos2 t tan2 t = 1 − cos2 t
cos2 t tan2 t + cos2 t = 1
cos2 t(tan2 t + 1) = 1
cos2 t =
1
tan t + 1
2
cos t = ± √
1
tan2 t + 1
Since cos t is negative in Quadrant II, the negative sign applies here. Thus
1
cos t = − √ 2
tan t + 1
9