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(6 – 1)
Angle and their Measure
Learning target: To convert between decimals and degrees, minutes, seconds forms
To find the arc length of a circle
To convert from degrees to radians and from radians to degrees
To find the area of a sector of a circle
To find the linear speed of an object traveling in circular motion
Vocabulary: Initial side & terminal side:
Initial side


Initial side
Initial side
Drawing an angle
Positive angles: Counterclockwise
Negative angles: Clockwise
360
200
90
-90
150
-135
Another unit for an angle: Radian
Definition: An angle that has its vertex at the center of a circle and
that intercepts an arc on the circle equal in length to the radius of the
circle has a measure of one radian.
r

r
One radian
From Geometry:
C  2 r
The entire circle has 360
360  2 r
180   r or  r  180
r
180

Therefore: using the unit
circle r = 1
 = 180
So, one revolution 360 = 2
Converting from degrees to radians & from radians to degrees
Degrees
Radians
Degree multiply by 
Radians
Degrees
Radian multiply by 180

180

  180 
 
  60
3 3  
   5
150  150 
radians

 180  6
I do: Convert from degrees to radians or from radians to degrees.
(a) -45
(b)
3
2
You do: Convert from degrees to radians or from radians to degrees.
(a) 90
(c)  radians
4
(b) 270
(d) 3 radians
Special angles in degrees & in radians
degrees
0
30
45
60
90
180
360
0

6

4

3

2

2
Radians
Finding the arc length & the sector area of a circle
Arc length (s):

S

s  r
is the central angle.
r
Area of a sector (A):
Important:

is in radians.
1 2
A r 
2
(ex) A circle has radius 18.2 cm. Find the arc length and the area if
the central angle is 3/8.
Arc length:
s  r
 3 
s  (18.2)  
 8 
5406

 21.4cm
8
Area of the sector:
1 2
A r 
2
1
2  3 
A  (18.2)  
2
 8 
 3 
 (165.62)  
 8 
 195.1cm 2
You do: (ex) A circle has radius 18.2 cm. Find the arc length and the
area if the central angle is 144.
Arc length:
1. Convert the degrees to radians
2.
s  r
Area of the sector:
1 2
A r 
2
Trigonometric functions & Unit circle
Learning target: To find the values of the trigonometric functions using
a point on the unit circle
(6 – 2)
To find the exact values of the trig functions in different
quadrants
To find the exact values of special angles
To use a circle to find the trig functions
Vocabulary:
Unit circle is a circle with center at the origin and the radius of one
unit.
Unit circle
Recall: trig ratio from Geometry
Opposite
a
sin A 

Hypotenuse c
cos A 
adjacent
b

Hypotenuse c
opposite
tan A 
adjacent
SOH
CAH
TOA
Also, Two special triangles

3
30, 60, 90 triangle
2
2
1
60
1
90
30

6
3
3
45, 45, 90 triangle
2
2
45
1
1
45
90
1

4

4
1
Using the unit circle
y
sin  
r
r
y
x
cos  
r
x
y
tan  
x
Finding the values of trig functions
Now we have six trig ratios.
Opposite
y
sin  

h ypotenuse r
1
hypotenuse r
csc 


sin 
opposite
y
adjacent
x
cos  

h ypotenuse r
1
hypotenuse r
sec  


cos 
adjacent
x
opposite y
tan  

adjacent x
1
adjacent x
cot  


tan  opposite y
Sin is positive when  is in QI.

y 0
sin 0    0
r 1
sin 30  sin
sin 45  sin

6

4
sin 60  sin
sin 90  sin
=
=
Find the exact value of the trig ratios.

3

2

=
Sin is positive when  is in QII
y
sin   
r

2


0
2
sin120  sin

3
3
sin135  sin
4
3
2
5
sin150  sin
6
sin180  sin  
y 
sin     
r 
7
sin 210  sin

6
Sin is negativewhen  is in QIII
2


0
3
2
5
sin 225  sin
4
4
sin 240  sin
3
3
sin 270  sin
 1
2
Sin is negative when  is in QIV

y 
sin     
r 
5
sin 300  sin

3
7
sin 315  sin
4
11
sin 330  sin
6
sin 360  sin 2  1
x
cos   
r
cos is positive when  is
in QI
cos 30  cos

6
cos is negative when  is
in QII
3
cos135  cos
4
cos is negative when  is
in QIII
4
cos 240  cos
3
cos is positive when  is
in QIV
11
cos 330  cos
6
y
tan   
x
tan is positive when  is
in QI (+, +)

tan 60  tan
3
cos is negative when  is
in QII(-, +)
3
tan135  tan
4
cos is negative when  is
in QIII(-, -)
4
tan 240  tan
3
cos is positive when  is
in QIV(+, -)
11
tan 330  tan
6
Find the exact values of the trig ratios.
csc 60  csc

3
sec30  sec
cot 45  cot

6

4
11
csc 330  csc
6
5
sec 225  sec
4
cot 90  cot

2
csc180  csc 
sec90  sec

2
4
cot 240  cot
3
(6 – 3)
Properties of trigonometric functions
Learning target: To learn domain & range of the trig functions
To learn period of the trig functions
To learn even-odd-properties
Signs of trig functions in each quadrant
 in Q.
I
II
III
IV
sin
+
+
-
cos
+
+
tan
+
+
-
csc
+
+
-
sec
+
+
cot
+
+
-
(sin)(csc) = 1
(cos)(sec) = 1
cos 
cot  
(tan)(cot) = 1 sin 
sin 
tan  
cos 
cos 
cot  
sin 
The formula of a circle with
the center at the origin and the
radius 1 is:
x2  y 2  1
y y
sin     y
r 1
x x
cos     x
r 1
sin 2   cos2   y 2  x2  1
Therefore, sin 2   cos 2   1
Fundamental Identities:
(1) Reciprocal identities:
1
csc  
sin 
1
cot  
tan 
1
sec  
cos 
(2) Tangent & cotangent identities:
sin 
cos 
tan  
cot  
cos 
sin 
(3) Pythagorean identities:
sin   cos   1 tan   1  sec 
2
2
2
2
cot 2   1  csc 2 
Even-Odd Properties
sin( )   sin 
csc( )   csc 
cos( )  cos 
sec( )  sec 
tan( )   tan 
cot( )   cot 
Co-functions:
Find the period, domain, and range
y = sinx
• Period: 2
• Domain: All real
numbers
•Range:
-1  y  1
y = cosx
• Period: 2
• Domain: All real
numbers
•Range:
-1  y  1
y = tanx
• Period: 
• Domain: All real
number but
(2n  1)
x
2
•Range: - < y <
y = cotx
• Period: 
• Domain: All real
number but
x  n
•Range:
- < y <
y = cscx
y = cscx
y = sinx
• Period: 
• Domain: All real
number but
x  n
•Range: - < y  -1
or 1 y < 
y = secx
• Period: 
• Domain: All real
number but
(2n  1)
x
2
Range: - < y  -1
or 1 y < 
Summary for: period, domain, and range of trigonometric functions
Functions
y = sinx
Period Domain
2 All real #’s
Range
-1  y  1
y = cosx
2
All real #’s
-1  y  1
y = tanx

All real #’s but
(2n  1)
- < y <
x
2
y = cotx

All real #’s but
- < y <
y = cscx

All real #’s but
- < y  -1
or 1 y < 
y = secx

All real #’s but
(2n  1)
x
2
- < y  -1
or 1 y < 
x  n
x  n
(6 – 4)
Graph of sine and cosine functions
Learning target: To graph y = a sin (bx) & y = a cos (bx) functions
using transformations
To find amplitude and period of sinusoidal function
To graph sinusoidal functions using key points
To find an equation of sinusoidal graph
Sine function:
Notes: a function is defined as: y = a sin(bx – c) + d
Period : P 
Amplitude: a
2
b
Period and amplitude of y = sinx graph
 

2

2

3
2
2
5
2
Graphing a sin(bx – c) +d
a: amplitude = |a| is the maximum depth of the graph
above half and below half.
bx – c : shifting along x-axis
Set 0  bx – c  2 and solve for x to
find the starting and ending point of the graph
for 1 perid.
d: shifting along y-axis
2
Period: one cycle of the graph P 
b
I do (ex) : Find the period, amplitude, and sketch the graph y = 3
sin2x for 2 periods.
Step 1:a = |3|, b = 2, no vertical or horizontal shift
Step 2: Amplitude: |3|
Period:
2
P
b
Step 3: divide the period into 4 parts equally.
Step 4: mark one 4 points, and sketch the graph
y = 3 sin2x
a = |3|
P: 
3
-3
y = cos x 


2

2

3
2
2
5
2
Graphing a cos (bx – c) +d
a: amplitude = |a| is the maximum depth of the graph
above half and below half.
bx – c : shifting along x-axis
Set 0  bx – c  2 and solve for x to
find the starting and ending point of the graph
for 1 perid.
d: shifting along y-axis
2
Period: one cycle of the graph P 
b
We do: Find the period, amplitude, and sketch the graph
y = 2 cos(1/2)x for 1 periods.
Step 1:a = |2|, b = 1/2, no vertical or horizontal shift
Step 2: Amplitude: |2|
Period: P  2
b
Step 3: divide the period into 4 parts equally.
Step 4: mark the 4 points, and sketch the graph
2
-2
You do: Find the period, amplitude, and sketch the graph
y = 3 sin(1/2)x for 1 periods.
I do: Find the period, amplitude, translations, symmetric, and
sketch the graph
y = 2 cos(2x - ) - 3 for 1 period.
Step 1:a = |2|, b = 2
Step 2: Amplitude: |2|
Period: P  2
b
Step 3: shift the x-axis 3 units down.
Step 4: put 0  2x –   2 , and solve for x to find the beginning
point and the ending point.
Step 5: divide one period into 4 parts equally.
Step 6: mark the 4 points, and sketch the graph.
y = 2 cos(2x - ) – 3
a: |2|
Horizontal shift: /2  x  3/2,
P: 
Vertical shift: 3 units downward
We do: Find the period, amplitude, translations, symmetric, and
sketch the graph
y = -3 sin(2x - /2) for 1 period.
Step 1: graph y = 3 sin(2x - /2) first
Step 2:a = |3|, b = 2, no vertical shift
Step 3: Amplitude: |3|
2
Period: P 
b
Step 4: put 0  2x – /2  2 , and solve for x to find the beginning
point and ending point.
Step 5: divide one period into 4 parts equally.
Step 6: mark the 4 points, and sketch the graph with a dotted line.
Step 7: Start at -3 on the starting x-coordinates.
y = -3 sin(2x - /2)
a=3
P=
3
/4  x  5/4
No vertical shift
0
-3
You do: Find the period, amplitude, translations, symmetric, and
sketch the graph
y = 3 cos(/4)x + 2 for 1 period.
Step 1: graph y = 3 cos(/4)x first
Step 2:a = |3|, b = /4
Step 3: Shift 2 units upward
Step 4: Amplitude: |3|
2
Period: P 
b
Step 5: Step 5: divide one period into 4 parts equally.
Step 6: mark the 4 points, and sketch the graph with a dotted line.
(6 – 5)
Graphing tangent, cotangent, cosecant, and secant functions
Learning target: To graph functions of the form y = a tan(bx) + c
and y = a cot(bx) + c
To graph functions of the form y = a csc(bx) + c and
y = a sec(bx) + c
The graph of a tangent function
• Period: 
• Domain: All real
number but
(2n  1)
x
2
•Range: - < y <
interval: 

2
x

2
Tendency of y = a tan(x) graph
y = 2 tan(x)
y = tan(x)
y = ½ tan(x)
To graph y = a tan(bx + c):
(1)The period is 
b
and
c
(2) The phase shift is 
b
(3) To find vertical asymptotes for the graph: 
solve for x that shows the one period

2
 bx  c 

2
I do: Find the period and translation, and sketch the graph
y = ½ tan (x + /4)
a=½,
b = 1,
c = /4

P=
b


2
 x
Interval:

4


-3/4
2
3


x
4
4
One half of the interval is the zero point.
/4
We do: Find the period and translation, and sketch the graph

1
y   tan  x  
3
2
1

Graph y  tan  x  
3
2
a=1
b=½
c = /3

P= b
Interval:
- /2< (1/2)x + /3 < /2
first

1
y   tan  x  
3
2

1
y  tan  x  
3
2
a=1
P = 2
Interval:
-5/3 < x < /3
You do: Find the period and translation, and sketch the graph

y  tan( x  )
4
a=1
P=
Interval:
The graph of a cotangent function
• y = cot(x)
• Period: 
• interval:
x  n
0<x<
• Domain: All real
number but
x  n
•Range:
- < y <
The tendency of y = a cot(x)
1
y  cot( x)
2
As a gets smaller,
the graph gets
closer to the
asymptote.
y  cot( x)
y  2 cot( x)
Graphing cosecant functions
• Period: 
• Interval: 0 < x < 
• Domain: all real numbers,
but x  n
• Range: |y|  1 or
y  -1 or y  1
(-, -1]  [1, )
Step 1: y = cos(x), graph y = sin(x)
Step 2: draw asymptotes x-intercepts
Step 3: draw a parabola between each asymptote with the vertex at
y = 1
Graphing secant functions
• Period: 
• Interval: /2 < x < 3/2
• Domain: all real numbers,
but
(2n  1)
x
2
• Range: |y|  1 or
y  -1 or y  1
(-, -1]  [1, )
Graphing secant functions
Step 1: graph y = cos(x)
Step 2: draw asymptotes x-intercepts
Step 3: draw a parabola between each asymptote with the vertex at
y = 1
I do (ex) Find the period, interval, and asymptotes and sketch the
graph.
y  csc(2 x   )
Graph y = sin(2x - )
•Period: P = 2/|b|
• Interval: 0 <2x -  < 2
1
• draw the asymptotes
-1
•Draw a parabola between
the asymptotes
You do: Find the period, interval, and asymptotes and sketch the
graph.


y  sec  x  
2

Graph y = cos(x - /2)
•Period: P = 2/|b|
• Interval: 0 <x - /2 < 2
• draw the asymptotes
•Draw a parabola between
the asymptotes