Download Chapter 4 Angles and their measures

 What
do you know about degrees?
 What
do you know about Radians?
 Degree
is represented by the symbol °. It
is a unit of angular measure equal to
1/180th of a straight angle.
 In
DMS (Degrees-minutes-second)
system of angular measure, each degree
is subdivided into 60 minutes and each
minute is subdivided into 60 seconds.
 A)
Convert 37.425° 𝑖𝑛𝑡𝑜 𝐷𝑀𝑆
 B)
Convert 42°24′ 36′′ 𝑖𝑛𝑡𝑜 𝑑𝑒𝑔𝑟𝑒𝑒𝑠
 A)
First we convert .425° 𝑖𝑛𝑡𝑜 𝑚𝑖𝑛𝑢𝑡𝑒𝑠
60′
1°
• . 425° ∗
= 25.5′
• Then we have to convert .5 minute into seconds

. 5′
∗
60′′
1′
= 30′′
So the answer is 37°25′ 30′′
 Each
minute is 1/60th of a degree and
each second is 1/3600th of a degree
 So
it is 42° +
24
36
( )° + (
)°
60
3600
= 42.41°
In navigation, the course or bearing of an
object is sometimes given as the angle of
the line of travel measured clockwise
from due north.
Sli
de
4-
A central angle of a circle has measure 1
radian if it intercepts an arc with the
same length as the radius.
Sli
de
4-
180
.
 radians
 radians
To convert degrees to radians, multiply by
.
180
To convert radians to degrees, multiply by
Sli
de
4-
How many radians are in 60 degrees?
Sli
de
4-
𝜋

3
 How
 How
many radians are in 90 degrees?
many degrees are in
5𝜋
3
𝑟𝑎𝑑𝑖𝑎𝑛𝑠?
If  is a central angle in a circle of radius r , and if  is measured in
radians, then the length s of the intercepted arc is given by
s  r .
𝜃
You may seen this as 𝑠 = 360 ∗ 2𝜋𝑟
Sli
de
4-
If  is a central angle in a circle of radius r , and if  is measured in
degrees, then the length s of the intercepted arc is given by
s
 r
180
.
𝜃
You may seen this as 𝑠 = 360 ∗ 2𝜋𝑟
𝑤ℎ𝑒𝑛 𝑦𝑜𝑢 𝑠𝑖𝑚𝑝𝑙𝑖𝑓𝑦 𝑡ℎ𝑖𝑠 𝑖𝑡 𝑏𝑒𝑐𝑎𝑚𝑒 𝑡ℎ𝑒 𝑓𝑜𝑟𝑚𝑢𝑙𝑎 𝑜𝑛 𝑡ℎ𝑒 𝑡𝑜𝑝.
Sli
de
4-
 So
basically, when you find the arc length
of a circle, you are finding the radian!!
Find the perimeter of a 30 slice of a large 8 in. radius pizza.
Sli
de
4-
Find the perimeter of a 30 slice of a large 8 in. radius pizza.
Let s equal the arc length of the pizza's curved edge.
  8  30 
240
 4.2 in.
180
180
P  8 in.  8 in.  s in.
s

P  20.2 in.
Sli
de
4-
 Angular
speed is measured in units like
revolutions per minute.
 Linear speed is measured in units like
miles per hour.
Sli
de
4-
 You
have a car, it’s wheels are 36 inches
in diameter. If the wheels are rotating at
630 rpm, find the truck’s speed in miles
per hour
630 𝑟𝑒𝑣 60 𝑚𝑖𝑛 2𝜋 𝑟𝑎𝑑𝑖𝑎𝑛
18 𝑖𝑛
1 𝑓𝑡
1 𝑚𝑖𝑙𝑒

𝑥
𝑥
𝑥
𝑥
𝑥
1 𝑚𝑖𝑛
1 ℎ𝑜𝑢𝑟
1 𝑟𝑒𝑣
1 𝑟𝑎𝑑𝑖𝑎𝑛 12 𝑖𝑛 5280𝑓𝑡
 67.47
 Note
mi/hr
remember 1 radian = the radius
A nautical mile (naut mi) is the length of 1
minute of arc along Earth’s equator.
Sli
de
4-
 Statue
mile is the “land mile”
1 statute mile  0.87 nautical mile
1 nautical mile  1.15 statute mile
Sli
de
4-
 From
Boston to San Franciso is 2698 stat
mi, convert it to nautical mile.
 2698
𝑠𝑡𝑎𝑡 𝑚𝑖𝑙𝑒 ∗
.87
1 𝑠𝑡𝑎𝑡 𝑚𝑖𝑙𝑒
= 2345 𝑛𝑎𝑢𝑡 𝑚𝑖
P
356 #1-39 every other odd

Exploration activity!

You are to cut out the unit circle I provided onto the notebook.



You are to trace as many special right triangle onto the unit circle
as possible, but here is the rule. The hypotenuse of the triangle
must be the radius of the circle and one leg on the axis.
Hint: You should have 3 per quadrant
After tracing it all, find the coordinates of the points that lies right
on the circle and find the cumulative degrees of each point on the
circle. Then answer the following questions in your group
Slide 4- 29
 Why
is it called a unit circle?
 How
does the special right triangles and
unit circle relate? What does the special
right triangles give you relating to the
circle?
 Is
it possible to convert the degrees into
radians? How do you do it?
The unit circle is a circle of radius 1
centered at the origin.
Slide 4- 31
 SOHCAHTOA
𝑜𝑝𝑝
 𝑆𝑖𝑛𝜃 =
ℎ𝑦𝑝
𝑎𝑑𝑗
 𝐶𝑜𝑠𝜃 =
ℎ𝑦𝑝
𝑜𝑝𝑝
 𝑇𝑎𝑛𝜃 =
𝑎𝑑𝑗
 *Teacher
make up different problems
regarding SOHCAHTOA

Please remember the unit circle and the coordinates
with its radian/degrees


In an Unit Circle, the radius r is 1
𝜃 is the degree or radians

Also please note that (X,Y) is a rectangular
coordinate

In the unit circle with regarding to Trig:



(X,Y) = (𝑟𝐶𝑜𝑠 𝜃, 𝑟𝑆𝑖𝑛 𝜃)
X is r𝐶𝑜𝑠 𝜃
Y is r𝑆𝑖𝑛 𝜃
 *Teacher
make up different practices
regarding Unit Circle
𝑜𝑝𝑝
 sin 𝜃 =
ℎ𝑦𝑝
𝑎𝑑𝑗
 cos 𝜃 =
ℎ𝑦𝑝
𝑜𝑝𝑝
 t𝑎𝑛 𝜃 =
𝑎𝑑𝑗
1
ℎ𝑦𝑝
csc 𝜃 =
=
𝑆𝑖𝑛 𝜃
𝑜𝑝𝑝
1
ℎ𝑦𝑝
sec 𝜃 =
=
𝑐𝑜𝑠𝜃
𝑎𝑑𝑗
1
𝑎𝑑𝑗
cot 𝜃 =
=
tan 𝜃
𝑜𝑝𝑝
 Teacher
make up different practices
regarding the 6 trigonometry functions
Slide 4- 38
Slide 4- 39
Two angles in an extended anglemeasurement system can have the same
initial side and the same terminal side, yet
have different measures. Such angles are
called coterminal angles.
Slide 4- 40
Find a positive angle and a negative angle that are coterminal
with 45 .
Add 360 : 45  360  405
Subtract 360 : 45  360  315
Slide 4- 41
Find a positive angle and a negative angle that are coterminal
with

6
.
Slide 4- 42
Find a positive angle and a negative angle that are coterminal
with

6
.
Add 2 :

6
 2 
Subtract 2 :

6
13
6
 2  
11
6
Slide 4- 43
Let  be the acute angle in standard position whose terminal
side contains the point (3,5). Find the six trigonometric functions
of  .
Slide 4- 44
Let  be the acute angle in standard position whose terminal
side contains the point (3,5). Find the six trigonometric functions
of  .
The distance from (3,5) to the origin is 34.
sin  
5
cos  
3
tan  
34
34
5
3
 0.857
 0.514
csc  
34
 1.166
5
34
sec  
 1.944
3
3
cot  
5
Slide 4- 45
Find sin 210 without a calculator.
Slide 4- 46
Find sin  and cos  , given tan   4 / 3 and cos   0.
Slide 4- 47
P
366 #1-55 EOO
P
381 #1-48 EOE
Function
Inverse Function
𝑠𝑖𝑛 𝜃
sin−1 𝜃 𝑜𝑟 arcsin 𝜃
cos 𝜃
cos −1 𝜃 𝑜𝑟 arccos 𝜃
tan 𝜃
tan−1 𝜃 𝑜𝑟 arctan 𝜃
1
sin−1 𝜃
1
cos −1 𝜃
csc 𝜃
sec 𝜃
cot 𝜃
1
tan−1 𝜃
−1
 sin
𝜋 𝜋
[− , ]
2 2
𝜃=
−1
 cos
𝜃 = [0, 𝜋]
𝜋 𝜋
−1
 tan
𝜃 =[− , ]
2 2
 Why
is that? We will look at this when we
graph it.
 *Teacher
make up different problems
 Pg
421 #1-32 EOO, 47,54
 1)
Draw a coordinate plane
 2) Label x-axis "𝜃“
 3) Label y-axis “y”
 Then
graph 𝑦 = sin 𝜃 (shown in class)
 Students
graph in groups:
• 𝑦 = cos 𝜃 and 𝑦 = tan 𝜃
A function is a sinusoid if it can be written in the form
f ( x)  a sin(bx  c)  d where a, b, c, and d are constants
and neither a nor b is 0.
Slide 4- 57
The amplitude of the sinusoid f ( x)  a sin(bx  c)  d is |a|.
Similarly, the amplitude of f ( x)  a cos(bx  c)  d is |a|.
Graphically, the amplitude is half the height of the wave.
Slide 4- 58
The period of the sinusoid f ( x)  a sin(bx  c)  d is 2 / | b | .
Similarly, the period of f ( x)  a cos(bx  c)  d is 2 / | b | .
Graphically, the period is the length of one full cycle of the wave.
Slide 4- 59
x
Find the period of y  sin   and use the language of transformations
2
to describe how the graph relates to y  sin x.
Slide 4- 60
x
Find the period of y  sin   and use the language of transformations
2
to describe how the graph relates to y  sin x.
2
x
The period is
 4 . The graph of y  sin   is a horizontal
1
2
2
stretch of y  sin x by a factor of 2.
Slide 4- 61
The frequency of the sinusoid f ( x)  a sin(bx  c)  d is | b | / 2 .
Similarly, the frequency of f ( x)  a cos(bx  c)  d is | b | / 2 .
Graphically, the frequency is the number of complete cycles the
wave completes in a unit interval.
Slide 4- 62
Construct a sinusoid with period  /3 and amplitude 4 that goes through (2,0).
Slide 4- 63
Construct a sinusoid with period  /3 and amplitude 4 that goes through (2,0).
To find the coefficient of x, set 2 / | b |  / 3 and solve for b.
Find b  6. Arbitrarily choose b  6.
For the amplitude set | a | 4. Arbitrarily choose a  4.
The graph contains (2,0) so shift the function 2 units to the right.
y  4sin(6( x - 2))  4sin(6 x -12).
Slide 4- 64
The graphs of y  a sin(b( x  h))  k and y  a cos(b( x - h))  k (where a  0 and
b  0) have the following characteristics:
amplitude = |a | ;
2
;
|b|
|b|
frequency =
.
2
When complared to the graphs of y  a sin bx and y  a cos bx, respectively,
period =
they also have the following characteristics:
a phase shift of h;
a vertical translation of k .
Slide 4- 65
1. Determine the maximum value M and minimum value m. The amplitude A
M -m
M m
, and the vertical shift will be C 
.
2
2
2. Determine the period p, the time interval of a single cycle of the periodic
of the sunusoid will be A 
2
.
p
3. Choose an appropriate sinusoid based on behavior at some given time T .
function. The horizontal shrink (or stretch) will be B 
For example, at time T :
f (t )  A cos( B(t - T ))  C attains a maximum value;
f (t )  - A cos( B(t - T ))  C attains a minimum value;
f (t )  A sin( B (t - T ))  C is halfway between a minimum and a maximum value;
f (t )  - A sin( B(t - T ))  C is halfway between a maximum and a minimum value.
Slide 4- 66
P
392 #1-52 EOE
 Why
are there vertical asymptotes?
 Why
are there vertical asymptotes?
 Graph
y=cos x
 Graph
y = sin x
 sec 𝑥
=2
 csc 𝑥 = 2
 cot 𝑥 = − 3
 sec 𝑥 = − 2
 csc 𝑥 = 1
 cot 𝑥 = 1
 Pg
402 #17-34