Algebra 2 Interactive Chalkboard Copyright © by The McGraw-Hill Companies, Inc. Send all inquiries to: GLENCOE DIVISION Glencoe/McGraw-Hill 8787 Orion Place Columbus, Ohio 43240
Download ReportTranscript Algebra 2 Interactive Chalkboard Copyright © by The McGraw-Hill Companies, Inc. Send all inquiries to: GLENCOE DIVISION Glencoe/McGraw-Hill 8787 Orion Place Columbus, Ohio 43240
Algebra 2 Interactive Chalkboard
Copyright © by The McGraw-Hill Companies, Inc.
Send all inquiries to:
GLENCOE DIVISION Glencoe/McGraw-Hill 8787 Orion Place Columbus, Ohio 43240
Graphing Trigonometric Functions
Translations of Trigonometric Graphs
Verifying Trigonometric Identities
Sum and Difference of Angles Formulas
Double-Angle and Half-Angle Formulas
Solving Trigonometric Equations
Example 1 Graph Trigonometric Functions
Example 2 Use Trigonometric Functions
Find the amplitude and period for Then graph the function.
First, find the amplitude.
The coefficient of Next, find the period.
.
Use the amplitude and period to graph the function.
Answer:
amplitude:
1
; period:
1080
or
6
Find the amplitude and period for Then graph the function.
Amplitude:
.
Period:
Answer:
amplitude: period:
360
or
2
Find the amplitude and period for Then graph the function.
Amplitude: Period:
.
Answer:
amplitude:
2
; period:
1440
or
8
Find the amplitude and period for each function.
Then graph the function.
a.
Answer:
amplitude:
1
; period:
720
or
4
b.
Answer:
amplitude: period:
360
or
2
c.
Answer:
amplitude:
3
; period:
720
or
4
Oceanography The tidal range in the Bay of Fundy in Canada measures 50 feet. A tide is at equilibrium when it is at its normal level, halfway between its highest and lowest points. Write a function to represent the height
h
of the tide. Assume that the tide is at equilibrium at
t
= 0
and that the high tide is beginning.
Since the height of the tide is
0
function
h
=
a
sin
bt
, where
a
at
t
= 0
, use the sine is the amplitude of the tide and
t
is the time in hours.
Find the amplitude. The difference between high tide and low tide is the tidal range or 50 feet.
Find the value of
b
. Each cycle lasts about 12 hours.
Solve for
b
.
Answer:
Thus, an equation to represent the height of the tide is
Graph the tide function.
Answer:
Oceanography The tidal range of a body of water measures 28 feet. A tide is at equilibrium when it is at its normal level, halfway between its highest and lowest points. a.
Write a function to represent the height
h
of the tide. Assume that the tide is at equilibrium at
t
= 0
and that the high tide is beginning.
Answer:
b.
Graph the tide function.
Answer:
Example 1 Graph Horizontal Translations
Example 2 Graph Vertical Translations
Example 3 Graph Transformations
Example 4 Use Translations to Solve a Problem
State the amplitude, period, and phase shift for . Then graph the function.
Since
h a
= 2
and
b
= 1
, the amplitude and period of the function are the same as
y
= 2 cos
. However
= –20
, so the phase shift is
–20
. Because
h
< 0
, the parent graph is shifted to the left.
To graph
y y
= 2 sin 20
= 2 sin (
to the left.
+ 20
)
, consider the graph of . Graph this function and then shift the graph
Answer:
amplitude:
2
; period:
360
; phase shift:
20
left
State the amplitude, period, and phase shift for Then graph the function.
Amplitude: Period: Phase shift:
The phase shift is to the right, since
h
> 0
.
Answer:
amplitude: period:
2
; phase shift: right
State the amplitude, period, and phase shift for each function. Then graph the function.
a.
Answer:
amplitude:
3
; period:
360
; phase shift:
–30
b.
Answer:
amplitude: period:
2
; phase shift:
State the vertical shift, equation of the midline, amplitude, and period for . Then graph the function.
Since vertical shift is
–1
. Draw the midline,
y
amplitude is
2
and the period is
2
. and the
= –1
. The Draw the graph of the function relative to the midline.
Answer:
vertical shift:
–1
; midline:
y
amplitude:
= –1
;
2
; period:
2
State the vertical shift, equation of the midline, amplitude, and period for graph the function.
. Then
Vertical shift:
k
= 3
, so the midline is the graph of
y
= 3
.
Amplitude: Period:
Since the amplitude of the function is draw dashed lines parallel to the midline that are unit above and below the midline. Then draw the cosine curve.
Answer:
vertical shift:
+3
; midline:
y
= 3
; amplitude: period:
2
State the vertical shift, equation of the midline, amplitude, and period for each function. Then graph the function.
a.
Answer:
vertical shift:
–2
; midline:
y
= –2
; amplitude:
3
; period:
2
State the vertical shift, equation of the midline, amplitude, and period for each function. Then graph the function.
b.
Answer:
vertical shift:
2
; midline:
y
= 2
; amplitude:
3
; period:
2
State the vertical shift, amplitude, period, and phase shift of the function.
The function is written in the form Identify the values of
k
,
a
,
b
, and
h
.
so the vertical shift is
4
.
so the amplitude is or
3
.
Then graph
so the period is so the phase shift is right.
Graph the function.
Step 1
The vertical shift is
4
. Graph the midline
y
= 4
.
Step 2
The amplitude is
3
. Draw dashed lines
3
above and below the midline at
y
= 1
units and
y
= 7
.
Step 3
The period is , so the graph is compressed. Graph using the midline as a reference.
Step 4
Shift the graph to the right.
State the vertical shift, amplitude, period, and phase shift of the function.
Then graph Answer:
vertical shift:
–2
; amplitude:
2
; period: phase shift:
Answer:
Health Suppose a person’s resting blood pressure is 120 over 90. This means that the blood pressure oscillates between a maximum of 120 and a minimum of 90. Write a sine function that represents the blood pressure for
t
seconds if this person’s resting heart rate is 75 beats per minute. Then graph the function. Explore
You know that the function is periodic and can be modeled using sine.
Plan
Let
P
represent blood pressure and
t
represent time in seconds. Use the equation
Solve •
Write the equation for the midline. Since the maximum is
120
and the minimum is
90
, the midline lies halfway between these values.
•
Determine the amplitude by finding the difference between the midline value and the maximum and minimum values.
Thus, .
•
Determine the period of the function and solve for
b
. Recall that the period of a function can be found using the expression Since the heart rate is
75
beats per minute, there are
5
heartbeats every
4
seconds. So, the period is second. Write an equation.
Multiply by
5
and Solve.
For this example, let The use of the positive or negative value depends upon whether you begin a cycle with a maximum value (positive) or a minimum value (negative).
•
There is no phase shift, so
h
= 0
.
Answer:
The equation is
Graph the function.
Step 1
Draw the midline
P
= 105
.
Step 2
Draw the maximum and minimum reference lines.
Step 3
Use the period to draw the graph of the function.
Step 4
There is no phase shift.
Examine
Notice that each cycle begins at the midline, rises to
120
, drops to
90
, and then returns to the midline. This represents the blood pressure of
120
over
90
for one heartbeat. Since each cycle lasts second, there will be
5
heartbeats every
4
heartbeats in
1
seconds or
75
minute. Therefore, the graph accurately represents the information.
Health Write a sine function that represents the blood pressure for
t
seconds of a person with a resting blood pressure of 140 over 90 and a resting heart rate of 60 beats per minute. Then graph this function.
Answer:
Example 1 Find a Value of a Trigonometric Function
Example 2 Simplify an Expression
Example 3 Simplify and Use an Expression
Find
tan
if
sec
= –2
and
180
<
< 270
.
Trigonometric identity Subtract
1
from each side.
Substitute
–2
for
sec
.
Square
–2
.
Subtract.
Answer:
Take the square root of each side.
Since is in the third quadrant,
tan
positive. Thus, is
Find
sin
if
cos
=
and
90
<
< 180
.
Trigonometric identity Substitute for
cos
.
Square Subtract.
Take the square root of each side.
Answer:
Since is in the third quadrant,
tan
is positive. Thus,
a. Find
cos
if
sin
=
and
0
<
< 90
.
Answer: b. Find
sec
if
tan
= 2
and
0
<
< 90
.
Answer:
Simplify Answer:
Distributive Property Simplify.
Simplify Answer:
1
Baseball A model for the height of a baseball after it is hit as a function of time can be determined using trigonometry.
If the ball is hit with an initial velocity of
v
second at an angle of
feet per from the horizontal, then the height
h
of the ball after
t
seconds can be represented by where
h
0
is the height of the ball in feet the moment it is hit. Rewrite the equation in terms of
sec
.
Original equation Factor.
Answer:
Thus,
Baseball A formula for the height
h
of a baseball after is hit is where
is the measure of the angle of elevation of the initial path of the ball, object, and
g v
0
is the initial velocity of the is the acceleration due to gravity. Rewrite the equation in terms of
sin
. Answer:
Example 1 Transform One Side of an Equation
Example 2 Find an Equivalent Expression
Example 3 Verify by Transforming Both Sides
Verify that Answer:
Transform the left side.
is an identity.
Original equation Simplify.
Verify that Answer: is an identity.
Multiple-Choice Test Item A B C
0
D Read the Test Item
Find an expression that is equal to the given expression.
Solve the Test Item
Write a trigonometric identity by using the basic trigonometric identities and the definitions of trigonometric functions to transform the given expression to match one of the choices.
Rewrite using the LCD,
Subtract.
Simplify.
Answer:
Since the answer is A.
Multiple-Choice Test Item A C
0
Answer:
B
B D
Verify that Answer: is an identity.
Original equation Express all terms using sine and cosine.
Rewrite using the LCD, Simplify the right side.
Verify that Answer: is an identity.
Example 1 Use Sum and Difference of Angles Formulas
Example 2 Use Sum and Difference Formulas to Solve a Problem
Find the exact value of
sin 75
.
Use the formula Sum of angles Evaluate each expression.
.
Answer:
Multiply.
Simplify.
Find the exact value of
cos (–75
)
.
Use the formula Difference of angles Evaluate each expression.
Answer:
Multiply.
Simplify.
Find the exact value of each expression.
a.
sin 105
Answer: b.
cos (–120
)
Answer:
Physics On June 22, the maximum amount of light energy falling on a square foot of ground at a location in the northern hemisphere is given by where
is the latitude of the location and
E
is the amount of light energy when the Sun is directly overhead. Use the difference of angles formula to determine the amount of light energy in Raleigh, North Carolina, located at
35.8
N
.
Use the difference formula for sine.
Answer:
In Raleigh, North Carolina, the maximum light energy per square foot is
0.9770
E
.
Use the formula and the difference of angles formula to determine the amount of light energy in Columbus, Ohio, located at
40
N
.
Answer:
0.9588
E
Verify that Answer: is an identity.
Original equation Difference of angles formula Evaluate each expression.
Simplify.
Verify that Answer: is an identity.
Original equation Difference of angles formula Evaluate each expression.
Simplify.
Verify that each of the following is an identity.
a.
Answer:
Verify that each of the following is an identity.
b.
Answer:
Example 1 Double-Angle Formulas
Example 3 Evaluate Using Half-Angle Formulas
Find the value of
Use the identity First find the value of
if and is between
Subtract.
Find the square root of each side.
Since is in the first quadrant, cosine is positive. Thus,
Now find
Answer:
The value of Double-angle formula Simplify.
Find the value of if and is between
Double-angle formula
Answer:
The value of
cos 2
Simplify.
Find the value of each expression if is between a.
Answer: b.
Answer: and
Find second quadrant.
Since
is in the
we must find first.
Simplify.
Since is in the second quadrant, Take the square root of each side. Half-angle formula
Simplify the radicand.
Rationalize.
Multiply.
Answer:
Since is between Thus, positive and equals is
Find Answer: is in the fourth quadrant.
Find the exact value of half-angle formulas.
by using the
Answer:
Simplify the radicand.
Simplify the denominator.
Find the exact value of half-angle formulas.
by using the
Simplify the radicand.
Simplify the denominator.
Answer:
Since is in the third quadrant, is negative. Thus,
Find the exact value of each expression by using the half-angle formulas.
a.
Answer: b.
Answer:
Verify that an identity.
Answer: is
Original equation Distributive Property Simplify.
Multiply.
Verify that is an identity.
Answer:
Example 1 Solve Equations for a Given Interval
Example 2 Solve Trigonometric Equations
Example 3 Solve Trigonometric Equations Using Identities
Example 4 Determine Whether a Solution Exists
Example 5 Use a Trigonometric Equation
Find all solutions of
0
<
360
.
for the interval
Original equation Solve for
0
.
Distributive Property Simplify.
Divide each side by
–1
.
Factor.
Now use the Zero Product Property.
or
Answer:
The solutions are
30°
,
150°
, and
270
° .
Find all solutions of
0
<
2
.
for the interval
Original equation Solve for
0
.
Factor
Use the Zero Product Property.
or
Answer:
The solutions are
Find all solutions of each equation for the given interval.
a.
Answer: b.
Answer:
Solve is measured in radians. for all values of if
or Original equation Subtract Factor.
Zero Product Property Solve.
Look at the graph of to find solutions of The solutions are so on.
and so on, and
The only solution in the interval
0
to The period of the sine function is are and radians. So the solutions can be written as and where
k
is any integer. Similarly, the solutions for
Answer:
The solutions are and
Solve measured in degrees. for all values of if is
Original equation Solve for
0
.
Factor.
Solve for in the interval of or
Answer:
The solutions are
a. Solve for all values of measured in radians.
Answer: if is b. Solve measured in degrees.
Answer: for all values of if is
Solve
Original equation Multiply.
Factor.
or
Check Answer:
The solution is is undefined.
Thus, is not a solution.
Solve Answer:
The solution is .
Solve
Original equation Subtract
1
and add to each side.
Multiply each side by
4
.
Factor.
or is undefined for
Answer:
The solutions are and
Solve Answer:
Gardening Rhonda wants to wait to plant her flowers until there are at least 15 hours of daylight. The number of hours of daylight
H
in her town can be represented by where
d
is the day of the year and angle measures are in radians. On what day is it safe for Rhonda to plant her flowers?
Original equation
Subtract
11.45
each side.
from Divide each side by
6.5
.
Add
1.333
each side.
to Divide each side by
0.0168
.
Answer:
Rhonda can plant her flowers at about the 114 th day of the year, or around April 24 th .
Gardening If Rhonda decides to wait until there are 16 hours of daylight, on what day can she plant her flowers? Use the formula where
H
is the number of hours of daylight and
d
the year.
is the day of Answer:
around the 126 th May 6th day of the year, or around
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