10.8
TRIGONOMETRIC FUNCTIONS AND THEIR DERIVATIVES
Fig. 10-3
Refer to Fig. 10-3. Place an arrow OA of unit length so that its initial point O is the origin of a coordinate
system and its endpoint A is (1,0). Rotate OA about the point O through an angle with radian measure 0. Let
OB be the final position of the arrow after the rotation. Then cos 6 is defined to be the ^-coordinate of B, and
sin 0 is defined to be the ^-coordinate of B.
10.9
State the values of cos 0 and sin 0 for 0 = 0, 77/6, ir/4, ir/3, ir!2, -IT, 3ir/2, 2ir, 9ir/4.
10.10
sin 6 and cos (6
V2/2.
Evaluate: (a)cos(-ir/6) (b) sin (-7T/6) (c) cos(27r/3) (d) sin (2ir/3)
(a)
In general,
cos (-0) = cos ft Hence,
cos (-ir/6) = cos (77/6) = V5/2.
(*)
In general,
sin(-0)= -sin ft Hence, sin(-ir/6) = -sin (ir/6) = -|. (c) 2ir/3 = ir/2 + ir/6. We use the identity
cos (0 + ir/2) = -sin ft Thus, cos(2ir/3)= -sin (-rr/6) = -\. (d) We use the identity sin (0 + ir/2) =
cos ft Thus, sin(27r/3) = cos(7r/6) = V3/2.
10.11
Sketch the graph of the cosine and sine functions.
We use the values calculated in Problem 10.9 to draw Fig. 10-4.
10.12
Sketch the graph of y = cos 3*.
Because cos 3(* + 2tr/3) = cos (3>x + 2ir) = cos 3x, the function is of period p = 2ir/3. Hence, the
length of each wave is 277/3. The number/of waves over an interval of length 2ir is 3. (In general, this number
/, called the frequency of the function, is given by the equation /= 2ir/p.) Thus, the graph is as indicated in
Fig. 10-5.
10.13
Sketch the graph of y = 1.5 sin 4*.
The period p = ir/2. (In general, p = 2ir/b, where b is the coefficient of x.) The coefficient 1.5 is the
amplitude, the greatest height above the x-axis reached by points of the graph. Thus, the graph looks like Fig.
10-6.
Give the definition of sin 0 and cos ft
Fig. 10-2
63
Notice that 9ir/4 = 27r+ ir/4, and the sine and cosine functions have a period of 2ir, that is, sin(fl + 2ir) =
and cos (97T/4) = cos (Tr/4) =
1- 277-) = cos «. Hence, sin(97r/4) = sin(7r/4) = V2/2
e
0
7T-/6
7T/4
IT/3
it 12
IT
37T/2
2lT
sin 6
0
1/2
V2/2
V3/2
1
0
-1
0
cos 0
1
V5/2
V2/2
1/2
0
-1
0
1
TRIGONOMETRIC FUNCTIONS AND THEIR DERIVATIVES
Fig. 10-3
Refer to Fig. 10-3. Place an arrow OA of unit length so that its initial point O is the origin of a coordinate
system and its endpoint A is (1,0). Rotate OA about the point O through an angle with radian measure 0. Let
OB be the final position of the arrow after the rotation. Then cos 6 is defined to be the ^-coordinate of B, and
sin 0 is defined to be the ^-coordinate of B.
10.9
State the values of cos 0 and sin 0 for 0 = 0, 77/6, ir/4, ir/3, ir!2, -IT, 3ir/2, 2ir, 9ir/4.
10.10
sin 6 and cos (6
V2/2.
Evaluate: (a)cos(-ir/6) (b) sin (-7T/6) (c) cos(27r/3) (d) sin (2ir/3)
(a)
In general,
cos (-0) = cos ft Hence,
cos (-ir/6) = cos (77/6) = V5/2.
(*)
In general,
sin(-0)= -sin ft Hence, sin(-ir/6) = -sin (ir/6) = -|. (c) 2ir/3 = ir/2 + ir/6. We use the identity
cos (0 + ir/2) = -sin ft Thus, cos(2ir/3)= -sin (-rr/6) = -\. (d) We use the identity sin (0 + ir/2) =
cos ft Thus, sin(27r/3) = cos(7r/6) = V3/2.
10.11
Sketch the graph of the cosine and sine functions.
We use the values calculated in Problem 10.9 to draw Fig. 10-4.
10.12
Sketch the graph of y = cos 3*.
Because cos 3(* + 2tr/3) = cos (3>x + 2ir) = cos 3x, the function is of period p = 2ir/3. Hence, the
length of each wave is 277/3. The number/of waves over an interval of length 2ir is 3. (In general, this number
/, called the frequency of the function, is given by the equation /= 2ir/p.) Thus, the graph is as indicated in
Fig. 10-5.
10.13
Sketch the graph of y = 1.5 sin 4*.
The period p = ir/2. (In general, p = 2ir/b, where b is the coefficient of x.) The coefficient 1.5 is the
amplitude, the greatest height above the x-axis reached by points of the graph. Thus, the graph looks like Fig.
10-6.
Give the definition of sin 0 and cos ft
Fig. 10-2
63
Notice that 9ir/4 = 27r+ ir/4, and the sine and cosine functions have a period of 2ir, that is, sin(fl + 2ir) =
and cos (97T/4) = cos (Tr/4) =
1- 277-) = cos «. Hence, sin(97r/4) = sin(7r/4) = V2/2
e
0
7T-/6
7T/4
IT/3
it 12
IT
37T/2
2lT
sin 6
0
1/2
V2/2
V3/2
1
0
-1
0
cos 0
1
V5/2
V2/2
1/2
0
-1
0
1