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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 negativewhen 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
sec 30 sec
cot 45 cot
6
4
11
csc 330 csc
6
5
sec 225 sec
4
cot 90 cot
2
csc180 csc
sec 90 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 csc2
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
- < y <
•Range:
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