Transcript Fig1.1 - I-Shou University

Chapter 8 Trigonometric Functions
8.1
8.2
8.3
8.4
Radian Measure of Angles
The Sine, Cosine, and Tangent
Derivatives of Trigonometric Functions
Integrals of Trigonometric Functions
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Radian Measure of Angles
There are two methods of measuring angles─by degrees and by radians.
In measuring by degrees, we arbitrarily divide the full circle into 360
equal segments.
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The angle  shown in Figure 8.1.1 cuts off of the circle, and we assign it 45 degrees,
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A more natural method is to use the length of the arc cut off by the angle on a circle of
radius 1. In Figure 8.1.2, s is the length of arc cut off by the angle  .
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Radian Measure of Angles
Definition 8.1.1
【數】弧度
The radian measure of an angle is the length of the arc cut off on a circle of
radius 1 by the sides of the angle with its vertex at the center of the circle.
360 degrees  2 radians,
1 degrees 

180
radians
and
1 radian 

180
degrees.
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Radian Measure of Angles
If the movement is in the counterclockwise direction, we assign a positive sign to
the radian measure. If it is in the clockwise direction, we assign a negative sign.
Figure 8.1.4 shows the cases of  4 and   4 radians.
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Radian Measure of Angles
The angles shown in Figure 8.1.5 illustrate a few simple examples.
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The Sine, Cosine, and Tangent
The sine, cosine, and tangent functions are defined in elementary trigonometry
by means of right triangles. We place an acute angle  at the base of a right
銳角
triangle, as shown in Figure 8.2.1.
Then we define the sine, cosine, and tangent of  as follows:
The Sine Function
sin  
O
.
H
The Cosine Function
co s  
A
.
H
The Tangent Function
tan  
O
.
A
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The Sine, Cosine, and Tangent
We can use right triangles to find the values of the
trigonometric functions for certain special angles.
(as shown n Figure 8.2.3)

1

1
, tan  1 .
, cos 
sin 
4
4
4
2
2

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The Sine, Cosine, and Tangent
If we draw a line from the top vertex perpendicular to the base,
as shown in Figure 8.2.4, the line divides the original triangle
into two right triangles, each with acute angles  3 and  6 .
sin

3

3
 1

, cos  , tan  3 .
2
3 2
3

1

3

1
.
sin  , cos 
, tan 
6 2
6
2
6
3
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The Sine, Cosine, and Tangent
The sine and cosine as coordinates of points on the circle
A more general way to define the trigonometric functions is to use a
circle of radius 1.
If ( x, y ) are the coordinates of the tip of the terminal side, as shown in
Figure 8.2.5, then
sin   y, co s   x, tan  
y
x
.
tan  
sin 
.
cos 
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The Sine, Cosine, and Tangent
Example 8.2.1
Find sin  , co s  , and tan  for   0,  ,  2, and 3 2.
Solution
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The Sine, Cosine, and Tangent
TABLE 8.2.1
(0,  2)

sin 
co s 
tan 
+
+
+
( 2,  )
( ,3 2)
(3 2, 2 )
+
-
+
+
-
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The Sine, Cosine, and Tangent
The trigonometric functions are periodic, with period 2 .In other words,
sin(2   )  sin  cos(2   )  cos  tan(2   )  tan  .
We see that if two angles have radian measures  and  respectively, their
terminal points are reflections of one another across the horizontal axis.
sin( )   sin  cos ( )  cos tan( )   tan  .
Another important property of the sin and cosine is the following:
sin 2   cos 2   1 .
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The Sine, Cosine, and Tangent
The graphs of sine, cosine, and tangent
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Derivatives of Trigonometric Functions
The derivatives of sin x and cos x are given by the following formulas:
d
sin x  cos x
dx
d
co s x   sin x.
dx
Example 8.3.1
Find
d
( x sin x)
dx
Solution
Using the product rule, we have
d
d
( x sin x)  x sin x  1 sin x  x cos x  sin x .
dx
dx
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Derivatives of Trigonometric Functions
Example 8.3.4
dy
Find if
dx
(i) y  cos2 x
and (ii) y  sin( x2  3x) .
Solution
(i) Using the chain rule with u  cos x and y  u 2 , we obtain
dy dy du

 2u sin x  2 cos x sin x.
dx du dx
(ii) Using the chain rule with u  x2  3x and y  sin u, we get
dy dy du

 (2u  3)co s u  (2 x  3) cos( x 2  3 x).
dx du dx
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Derivatives of Trigonometric Functions
The other trigonometric functions
There are three additional trigonometric functions, the secant, cosecant,
and cotangent, defined as follows:
The Secant Function
sec x 
1
.
cos x
c s cx 
1
.
sin x
The Cosecant Function
The Cotangent Function
1
cos x
cot x 

.
tan x sin x
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Derivatives of Trigonometric Functions
d
sec x
Example 8.3.6 Find
dx
Solution Using the chain rule, we obtain
d
d
1
1 d
sin x
sec x  (
)
cos x 
.
2
2
dx
dx cos x
cos x dx
cos x
We can rewrite the formula in the form
d
d
sec x  sec x tan x ,
csc x   csc x cot x ,
dx
dx
d
d
tan x  sec2 x ,
cot x   csc2 x .
dx
dx
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Integrals of Trigonometric Functions
From the derivative formulas (15) and (16) we get the
following antiderivative formulas:
 sin xdx   cos x  c
 cos xdx  sin x  c
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Integrals of Trigonometric Functions
Example 8.4.1
Find the area under the cosine curve between  2 and  2 (shown in Figure 8.4.1).
Solution
Using the antiderivative formula (41) and the fundamental theorem of calculus, we
get
 2


2
cos xdx  sin x
 2
 2
 sin


 sin (- )  2
2
2
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Integrals of Trigonometric Functions
Example 8.4.3 Find (i)  sin 3 t cos tdt , (ii)
2
t
cos(
t
 1)dt ,

(iii) 
 2
0
t cos tdt .
Solution
(i)Using the substitution u  sin t so that du dt  cos t , we get
u4
sin 4 t
 sin t cos tdt   u du  4  c  4  c
3
3
(ii)Using the substitution u  t 2  1, we have du  2tdt . Therefore
1
1
1
2
2
t
cos(
t

1))
dt

cos
udu

sin
u

c

sin(
t
 1)  c


2
2
2
(iii)We use integration by parts, with u  t and dv dt  cos t . Then du dt  1 and
v  sin t , so that

 2
0
t cos tdt  t sin t
 2
0

 2
0
sin tdt 

2
 cos t
 2
0


2
 1.
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Integrals of Trigonometric Functions
Example 8. 4. 4 Find
 tan xdx.
sin x
dx,
Solution Writing  tan xdx  
cos x
We make the substitution u  cos x, so that du dx   sin x. Therefore
1


xdx
tan
 u du   ln u  c   ln cos x  c.

So that
 tan xdx  ln sec x  c.
Similarly
 cot xdx   ln csc x  c.
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Integrals of Trigonometric Functions
Example 8. 4. 5 Find
 sec xdx
Solution We first change the form of the integrand by multiplying and dividing by the
same expression:
sec x  tan x
sec2 x  sec x tan x
 sec xdx   sec x sec x  tan x dx   sec x  tan x dx.
Letting u  tan x  sec x and du  (sec2 x  sec x tan x)dx, we obtain
 sec xdx  
Therefore,
du
 ln u  c.
u
+
 sec xdx  ln tan x  sec x  c.
Similarly
 csc xdx   ln cot x  csc x  c.
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