Download is in radians!!!!

Chapter 7
The Trigonometric Functions
1
Chapter 7 Topics
• Radian Measure
• Trig Functions of angles
• Evaluating Trig Functions
• Algebra and Trig Functions
• Right Triangle Trig
2
Chapter 7.1
Real Numbers
3
Overview
• Radians for Angle Measurement
• Radian Measure of Standard Angles
• Converting
4
Why Radians?
• Degrees are a bit artificial, radians refer to an actual measure
• Definition: If we put the vertex of an angle at the center of a
circle of radius r and let s be the arc length intercepted, then
the angle,  is defined as
 = s/r
• Because of the definition, angle is now related to a
measureable quantity, that is the ratio of two lengths, s and r
5
Radians
• Now, lets make some sense of this:
– The arc whose length is the entire circle is the
circumference
– The circumference is 2  r
– Therefore, the angle that encompasses the whole circle
is 2 , which in degree measure is 360
• Bottom line, the angle of a whole circle is 2 ;
from that you can determine all other angle measures
6
Therefore:
• Find the angle measure in radians of the following:
Degree
Radian
360
2
180
90
270
60
30
45
7
Angle Measures
Degree
Radian
360
2
180

90
/2
270
3/2
60
/3
30
/6
45
/4
8
Converting:
• Let d be an angle in degrees and  be its measure in radians
• 180 degrees is 
• Need to have 180/d be the same as  / , or
d/180 =  / 
• This is the same as /180 =  /d
•
9
Example
• Find 36 deg in radians:
36/180 = /
36/180 = 1/5 = 0.2
So  = 0.2 
• We will not ask you to do much converting, but you will need
to learn the common angles in radians
10
Angles in Degrees and Radians
Angle
0
30
45
60
90
120
135
150
180
Radian
0
/6
/4
/3
/2
2/3
3/4
5/6

Angle
180
210
225
240
270
300
315
330
360
Radian

7/6
5/4
4/3
3/2
5/3
7/4
11/6
2
11
Example, Finding Angle Measure
• Find the radian measure of the following angle:
• 120=
• -225=
• 270=
12
Solutions
• Find the radian measure of the following angle:
• 120= 2(60 ) = 2 (/3) = 2/3
• -225= - 5(45 ) = -5(/4) = -5/4 [= 2  - 5/4 = 3/4]
• 270= 3(90 ) = 3(/2)
13
More Examples
Find the following angles in radians
• 180 =
• 30  =
• 330  =
• -210  =
14
Solutions
Find the following angles in radians
• 180 =  rad
• 30  =/6 rad
• 330  =11 /6 rad
• -210  = -180-30 = -7 /6 rad
15
Example
Convert to radians:
-75 =
250  12’ =
16
Example (Solution)
Convert to radians:
-75 = -45  - 30  = - /4 - /6 = - 5/12
250  12’ = 250.2  = 259.2 ( /180 ) = 1.39 
17
Example (Solution)
Convert from radians to degrees
 / 24 =
-5 =
18
Example
Convert from radians to degrees
 / 24 =  /24 (180 / ) = 180  / 24 = 7.5 
-5 = -5 (180 / ) = 900  / 
19
More Examples
Convert to radians:
• 230 =
• -35  =
• 200  48’ =
20
Example (Solutions)
Convert to radians:
•230 = 23/18
•-35  =7 /6
•200  48’ = ? For this one, you need to convert 48’ to degrees.
48’ = 48/60= 0.8 degrees. Then do the multiplication:
(200.8/180) 
21
Convert to Degrees:
• /4 =
• /2 =
• 5 /6 =
• 6=
22
Solutions
Convert to Degrees:
•/4 = 45
•/2 =90
•5 /6 =150
•6  = 1080
23
More Examples
Find the terminal side quadrant:
•  = 3.9
•  = 5.4
•  = -4.3
24
Solutions
Find the terminal side quadrant:
• = 3.9, Q 3
• = 5.4, Q 4
• = - 4.3, Q 2
25
Arc Length Example
If a circle has radius 8 cm, and the arc length is 18 cm, what is
the radian measure of an angle ?
 = s/r = 18/8 = 9/4 = 2.25 rad
26
Some More Examples
Find the radian measure of the angle with arc s and
circle radius r
• s = 24 m, r = 4 m
• s = 10 ft, r = 10 ft
• s = 5.2 mm, r = 2.6 mm
• s = 11 cm, r = 20 m
27
Example (Solutions)
Find the radian measure of the angle with arc s and
circle radius r
• s = 24 m, r = 4 m;  = 6
• s = 10 ft, r = 10 ft;  = 1
• s = 5.2 mm, r = 2.6 mm;  = 2
• s = 11 cm, r = 20 m; equivalent to 11 cm and 2000 cm:
 = 11/2000
28
Sector Area
• The area of a sector is the portion of a circle with angle  and
radius r
– A whole circle has area  r2
– The area of a sector of angle  is r2/2
• Think about it. This formula gives the following for a circle of
radius 1:
– Angle , half circle, area =  r2/2
– Angle 2 , whole circle, area =  r2
– Angle /4, 1/8 of a circle, area =  r2/8
29
Summary
• We can convert from radians to degrees and vice versa: The
key is that 360 degrees is 2  radians
• Arc length is r , where  is in radians
• Sector area is r2/2
30
Angular Velocity
Angular velocity equals the amount of rotation per unit time
• Often designated as  (omega)
•  =  / t, the angle per time
For example, a Ferris wheel rotating at 10 revolutions per minute
has angular velocity of
=/t
 = 10 revolutions/min
each revolution is 2  rad
 = 10 (2 ) / 1 min = 20  rad/min
31
Linear Velocity
Linear velocity is distance traveled (or change in position)
per unit time
D = r t, r = D / t
For angular motion, the distance D is the length of the arc, s
rate = v = s / t
Since s = r , v = r  / t = r ( / t ) = r 
Linear velocity v = r 
32
Example
A point P rotates around the circumference of a circle with radius
r = 2 ft at a constant rate. If it takes 5 sec to rotate through an
angle of 510,
a. What is the angular velocity of P
b. What is the linear velocity of P
33
Solution
a.
510 = 510 ( / 180 )
= (510/180)  = 17/6 
 =  / t = (17/6) / 5 = (17/30) rad/sec
b.
V = r  = 2ft (17/30 ) = (17/15) ft/ sec
34
More Examples
Point P passes through angle  in time t as it travels around the
circle. Find its angular velocity in radians
• = 540 , t = 9 yr
• = 270 , t = 12 min
• = 690 , t = 5 sec
• = 300 , t = 5 hr
35
Solutions
Covert to radians,  = angle / time
• = 540 , t = 9 yr; 540  = 540 / 180 = 3 ,  = /3 rad/yr
• = 270 , t = 12 min;  = /8 rad/min
• = 690 , t = 5 sec;  = 23  / 30 rad/sec
• = 300 , t = 5 hr;  = /3 rad/hr
36
Examples
Point P travels around a circle of radius r, find its linear velocity
1.  = 12 rad/min, r = 15 ft
2.  = 2312  rad/sec, r = 0.01 km
3.  = 282, t = 4.1 min, r = 1.2 yd
4.  = 45, t = 3 hr, r = 2 mi
37
Solutions
V=r
1. 180 ft/min
2. 72.6 mph
3. 1.44 yd/min
4. 30 mph
38
Example
Using Angular Velocity to Determine Linear Velocity
The wheels on a bicycle have a radius of 13 in. How fast, mph, is
the cyclist going if the wheels turn at 300 rpm?
 = 300 rev/min = 300 (2  ) / min = 600  /min
V = r  = 13 in (600  /min)
1 mile = 5280 ft x 12 in /ft, 1 hour = 60 min
V = 13 in (600  /min) (60 min/hr) ( 1 mi / (5280 x 12 in)
V = 23.2 mph
39
Examples
• A belt goes around two wheels
• The radius of the smaller is 5cm and the larger 15 cm
• The angular speed of the larger wheel is 1800 RPM
• Find the angular speed of the larger in radians, the linear
speed of a point on the edge of the larger, the angular speed
of the smaller, and the angular speed of the smaller in rpm
40
Solution
• The radius of the smaller is 5cm and the larger 15 cm
• The angular speed of the larger wheel is 1800 RPM
• Find the angular speed of the larger in radians, the linear
speed of a point on the edge of the larger, the angular speed
of the smaller, and the angular speed of the smaller in rpm
• 3000 pi rad/min
• 54000 pi cm/min
• 10800 pi cm/min; the linear velocity has to be the same; v=r.
54000/5 = 10800
• Rpm = /2 pi, so10800 pi/2 pi = 5400 rpm
41
Summary
• Arc Length: s = r ,  in radians
• Angular Velocity:  =  / t,  in radians
• Linear Velocity: v = r 
• Area of a circular sector subtended by angle  :  r2/2,
 in radians
All assume  is in radians!!!!
42
Chapter 7.2
Trig Functions of Angles Using Radians
43
Overview
• Locate points on a unit circle
• Use special triangles to find points on a unit circle
• Define the 6 trig functions in in terms of points on the
unit circle
44
Unit Circle
• Circle of radius 1
• Center at the origin
(-1,0)
y
(0,1)
x
(0,0)
(1,0)
(0,1)
45
Exercise
• Find a point of the unit circle if y = 1/2
46
Solution
x2 + y2 = 1
x = ± sqrt (1 – ¼) = ± sqrt(3)/2
47
Symmetry
• Find the quadrant containing (-3/5, -4/5) and verify it is on the
unit circle
48
Solution
• Quadrant : x and y < 0, so is in quadrant 3
• Check: x2 + y2 = 1
(3/5) 2 + (4/5) 2 = (9 + 16)/25 = 1
49
More on Symmetry
• If the point (a,b) is on the unit circle, so are
 (-a, b)
 (a, -b)
 (-a, -b)
50
Special Triangles
Find points on the unit circle associated with /4: /4: /2 triangle
51
Solution
• In quadrant 1, the triangle has x and y values sqrt(2)/2
• In quadrant 2 it is (- sqrt(2)/2, sqrt(2)/2)
• In quadrant 3, both signs are –
• In quadrant 4, x is positive, y is negative
52
More Examples
• Do the same for a /6: /3: /4 triangle
53
Solutions
(± sqrt(3)/2, ± 1/2)
(± 1/2, ± sqrt(3)/2)
54
Points Associated with Rotation
Find the points on the unit circle associated with
a. 5/6
b. 4 /3
c. 7 /4
55
Solution
a. (-sqrt(3)/2, 1/2)
b. (-1/2, - sqrt(3)/2)
c. (sqrt(2)/2, -sqrt(2)/2)
56
Trig Functions
Find the six trig functions for  = 5  /4
57
Solution
Quadrant 3
• cos  = -1/sqrt(2)
• sin  = -1/sqrt(2)
• tan  = 1
• sec  = -sqrt(2)
• csc  = -sqrt(2)
• cot  = 1
58
Summary
• Can find points on the unit circle for angles in radians
• Can calculate trig functions for points on the unit circle
59
Chapter 7.3
Evaluating Trig Fucntions
60
Overview
• Define trig functions in terms of a real number t
• Find the number associated with special values of the
trig functions
• Find the real number t associated with any trig value
• In this section we cover reference angles
61
Real Numbers
• Integers … -2, -1, 0, 1, 2, …
• Rationals: any number that can be expressed as a fraction
Rationals include integers
• Reals: the non-imaginary numbers. Includes sqrt (2) and all
rational numbers
62
Trigonometry of Real Numbers
• Work with the unit circle: r=1.
if r = 1, then arc length s = r  = 
• That way, any function of  is a function of arc length, s
63
Reference Angle
• A reference angle is an acute angle associated with an angle
 formed by the x axis and the terminal ray of the angle. It is,
necessarily,  /2
64
Give the Reference Angles
• 70 deg
• 115 deg
• 200 deg
• 400 deg
• - 80 deg
• - 325 deg
65
Solution
• 70 deg; 70
• 115 deg; 65
• 200 deg; 20
• 400 deg; 40
• - 80 deg; 80
• - 325 deg; 55
66
Give the reference angle
• 4/3
• 5
• 5/6 
• 7/8 
67
Solution
• 4/3; /3
• 5 ; 
• 5/6 ; /6
• 7/8 ; /8
• 21/10 ; /10
68
Examples
Give the six trig functions of:
a. 11  / 6
b. 3 / 2
69
Solution
a. Quadrant 4:
cos = sqrt(3)/2
sin = -1/2
tan = -1/sqrt(3)
b. Along y axis
cos = 0
sin = -1
tan = undefined (-)
Given these, can find csc, sec, cot
70
More Examples
• csc (/6) =
• csc (5/6) =
• csc (11/6) =
• csc (- /6) =
• csc (-17/6) =
71
Solutions
• csc (/6) = 2
• csc (5/6) = 2 ; /6 away from 180
• csc (11/6) = - 2 ; /6 away from 360
• csc (- /6) = - 2 ; /6 away from 0
• csc (-17/6) = -2 ; /6 away from -18  /6 = -3  = -180
72
Examples
• cot  =
• cot 0 =
• cot /2 =
• cot 3 /2
73
Solutions
• cot  = undefined
• cot 0 =undefined
• cot /2 = 0
• cot 3 /2 = 0
74
Special Angles
Find t such that
a. cos t = -1/sqrt(2)
in quadrant 2
b. tan t = sqrt (3)
in quadrant 3
75
Solutions
Find t such that
a. cos t = -1/sqrt(2); Q2 (like a 45 deg angle)
t = 3 /4
b. tan t = sqrt (3); Q3(like a 60 deg angle)
t = 4  /3
76
What you need to know
• Determine any reference angle
• Know how to find special angles on the unit circle
• Calculate the trig functions of these special angles
77
Chapter 7.4
Algebra and the Trig Functions
78
Notation
• sin of the quantity a is written sin a or sin (a); parentheses are
not required
• When the argument of sin is an expression parentheses are
required: sin of the quantity (a + b) is written sin(a+b)
• The expression sin2a means (sin a)(sin a)
• The expression sin2 (a+b) means (sin(a+b))(sin(a+b))
• The expression sin (a+b) 2 means sin of the quantity (a+b)2 ,
the function is not squared, only its argument
79
Manipulating Trig Functions
• Add sin a and 1/cos a
• Simplify (Sec a + 1)/ (sin + tan)
80
Solution
• Add sin a and 1/cos a
find a common denominator: (sin a cos a + 1)/cos a
• Simplify (Sec a + 1)/ (sin a+ tan a)
Replace sec and tan by sin and cos
(1/cos a + 1) / (sin a + sin a / cos a)
Multiply top and bottom by cos a
(1 + cos a)/(sin a cos a + sin a)
This becomes
1+cos 𝑎
(sin 𝑎)(1+cos 𝑎)
= 1/sin a or csc a
81
Using identities to calculate functions
• Sin a = 2/3 in the second 2nd quad, find cos a and tan a
82
Solution
• Sin a = 2/3 in the second 2nd quad, find cos a and tan a
• 1 – sin2 a = cos2 a = 1-4/9 = 5/9
• Cos2 a = 5/9 or cos a = 5 /3
83
Example
• Simplify sin a/cos a + cos a/sin a
84
Solution
• Simplify sin a/cos a + cos a/sin a
• Get a common denominator, sin a cos a
• (sin2 a + cos2 a )/(sin a cos a) = 1/(sin a cos a)
85
Example
• cos a/(1-sin a) = (1+sin a)/cos a
86
Solution
• cos a/(1-sin a) = (1+sin a)/cos a
• Multiply both sides by cos a (1 – sin a) to remove the
denominators
• cos2 a = 1 – sin2 a which is the Pythagorean identity
• Note, in this case we manipulated both sides of the equation.
We could have worked with just one.
87
Examples
• 1/sin a – sin a = cot a cos a
88
Solution
• 1/sin a – sin a = cot a cos a
• Get rid of cot and get the left a common denominator
• (1 – sin2 a)/sin a = cos2 a / sin a
• Holds by Pythagorean identity
89
Example
• (cos a – sin a)2 + 2sin a cos a = 1
90
Solution
• (cos a – sin a)2 + 2sin a cos a = 1
• Multiply out parentheses
• Cos2 a + sin2 a – 2 sin a cos a + 2 sin a cos a = 1
• By Pythagoras becomes:
• 1+0=1
91
Example
• Tan a tan b = (tan a + tan b)/(cot a + cot b)
92
Solution
• Tan a tan b = (tan a + tan b)/(cot a + cot b)
• Eliminate the denominator
tan a tan b(cot a + cot b) = tan a + tan b
• Multiply left side out, remembering tan a cot a = 1
the left becomes tan b + tan a, which is the right
93
How to prove isn’t identity
• Show cos A + cos B = cos (A+B) is not an identity
• To prove something isn’t an identity, only need to show that it
doesn’t work for one value!
94
Solution
• Show cos A + cos B = cos (A+B) is not an identity
• Try A = B = 60 deg, cos 60 = ½
• Does ½ + ½ =1 = cos 120? Cos 120 is – cos 60 = - ½ NO
95
Which is an identity?
• (Csc a – cot a) = (1 + cos a)(1-cos a)
• Cot a + sin a/(1+cos a) = csc a
96
Solution
• (Csc a – cot a) = (1 + cos a)(1-cos a)
The right side becomes 1 – cos2a = 1,
the left is (1- cos a)/sin a; there are many values for
which that is not = 1; Not an identity
• Cot a + sin a/(1+cos a) = csc a
Simplify by multiplying by sin a
cos a + sin2 a/(1 + cos a) = 1
• Replace sin2 a by (1-cos2 a)
• We have cos a + 1 – cos a on the left which = 1
97
What you need to know
• Notation: sin2 a vs sin a2
• Need to be comfortable in manipulating trig functions to prove
identities
• Need to be able to disapprove identities by substituting a
value
98
Chapter 7.5
Right Angle Trigonometry
99
• We have already completed most of this section
• The section uses a given trig value to find the remainder,
e.g., if sin = 3/5, find cos and tan
100
Cofunctions
• Sin (90 – A) = cos A; If two acute angles, A and B, are
complementary, then sin A = cos B
• We have the three angles of the triangle as A, B, 90
• Clearly 90 – A = B if they are complementary
B
c
A
a
b
• From the picture, sin A = a/c = cos B, etc.
• Additionally, cos (90 – A ) = sin B
101
Using Cofunctions
• Given cos (75) = a find sin (15)
102
Solution
• Given cos (75) = a find sin (15)
• We know that cos (90 – x) = sin x, so sin (15) = a
103
Trig values in other quadrants
• We have done these
104
More Examples
105
• Give the reference angle for:
 7 /6
 24  /3
106
Solution
• Give the reference angle for:
 7 /6 :  / 6
 24  /3: 8  or 0
107
Find
• Sin /3 =
• Cos 2 /3 =
• Tan /3 =
• Cos 7 /4 =
• Tan 7/4 =
108
Solution
109
• Verify that (1/3, - 2 sqrt(2) / 3) is a point on the unit circle.
• Find the value of all the trig functions associated with
this point
110
Solution
111
• A camera crew rids a cart on a circular arc. The radius of the
arc is 75 ft and can sweep an angle of 172.5º in 20 sec.
– Find the length of the track in feet
– Find the angular velocity of the cart
– Find the linear velocity of the cart.
112
Solution
• A camera crew rids a cart on a circular arc. The radius of the
arc is 75 ft and can sweep an angle of 172.5º in 20 sec.
– Find the length of the track in feet
• S=r  = 75 ft (172.5)(/180) = 225.8 ft
– Find the angular velocity of the cart
• ω = /time = 172.5(/180) /20 = 0.15 rad/sec
– Find the linear velocity of the cart
• V = r ω = (0.15 rad/sec) (75ft) = 11.29 ft/sec
113
Example
• Find t, between 0 and 2  if
 Sin t = -1/2 in Q3
 Sec t = 2 sqrt(3)/3 in Q4
 Tan t = -1 in Q2
114
Solution
• Find t, between 0 and 2  if
 Sin t = -1/2 in Q3: t= 7 /6
 Sec t = 2 sqrt(3)/3 in Q4: t = 11 /6
 Tan t = -1 in Q2: t = 3 /4
115
Example
• If (20/29, 21/29) is a point on the central unit circle, use
symmetry to find 3 other points on the circle.
116
Solution
• If (20/29, 21/29) is a point on the central unit circle, use
symmetry to find 3 other points on the circle.
(-20/29, 21/29)
(-20/29, -21/29)
(20/29, -21/29)
117
• Convert to radians:
 300
 -7239’
 
118
• Convert to radians:
 300: 5  /3
 -7239’= -1.26; First change the minutes to seconds
  = 0.0548
119
• Convert to degrees:
 9.29
 -3  /2
 45
120
Solution
• Convert to degrees:
 9.29 = 532.3
 -3  /2 = -270
 45 = 2578.3
 Remember, to convert to degrees, multiply by 180 / 
121
Example
• Assume Memphis is in directly north of New Orleans, at
90 º W longitude. Find the distance between cities, in km, if
the radius of the earth is 6000 km, Memphis is at 35º north,
and New Orleans is at 29.6º.
122
Solution
• Assume Memphis is in directly north of New Orleans, at
90 º W longitude. Find the distance between cities, in km, if
the radius of the earth is 6000 km, Memphis is at 35º north,
and New Orleans is at 29.6º.
Angle is 35º - 29.6 º = 5.4º
5.4º (/180) = 0.094 rad
Radius is 6000 km,
so arclength is 6000 ( 0.094) = 565 km
123