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Algebra 2 Interactive Chalkboard
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GLENCOE DIVISION
Glencoe/McGraw-Hill
8787 Orion Place
Columbus, Ohio 43240
Lesson 14-1
Graphing Trigonometric Functions
Lesson 14-2
Translations of Trigonometric Graphs
Lesson 14-3
Trigonometric Identities
Lesson 14-4
Verifying Trigonometric Identities
Lesson 14-5
Sum and Difference of Angles Formulas
Lesson 14-6
Double-Angle and Half-Angle Formulas
Lesson 14-7
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 at t = 0, use the sine
function h = a sin bt, where a 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 a = 2 and b = 1, the amplitude and period of
the function are the same as y = 2 cos . However
h = –20, so the phase shift is –20. Because h < 0,
the parent graph is shifted to the left.
To graph y = 2 sin ( + 20), consider the graph of
y = 2 sin . Graph this function and then shift the graph
20 to the left.
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
and the
vertical shift is –1. Draw the midline, y = –1. The
amplitude is 2 and the period is 2.
Draw the graph of the
function relative to the
midline.
Answer: vertical shift: –1;
midline: y = –1;
amplitude: 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
above and below the midline.
Then draw the cosine curve.
Answer:
vertical shift: +3;
midline: y = 3;
amplitude:
period: 2
unit
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.
Then graph
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.
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 units
above and below the midline at y = 1 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 seconds or 75
heartbeats in 1 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.
Take the square root
of each side.
Answer: Since  is in the third quadrant, tan  is
positive. Thus,
Find sin  if cos  =
and 90 <  < 180.
Trigonometric identity
Substitute
Square
Subtract.
for cos .
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
Distributive
Property
Simplify.
Answer:
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 feet per
second at an angle of  from the horizontal, then
the height h of the ball after t seconds can be
represented by
where h0 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, v0 is the initial velocity of the
object, and g 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
is an identity.
Answer:
Transform the left side.
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
B
C 0
D
Answer: B
Verify that
is an identity.
Answer:
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
Example 3 Verify Identities
Find the exact value of sin 75.
Use the formula
.
Sum of
angles
Evaluate
each
expression.
Multiply.
Simplify.
Answer:
Find the exact value of cos (–75).
Use the formula
Difference of angles
Evaluate each
expression.
Multiply.
Simplify.
Answer:
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 0.9770E.
Use the formula
and the
difference of angles formula to determine the
amount of light energy in Columbus, Ohio,
located at 40 N.
Answer: 0.9588E
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 2 Half-Angle Formulas
Example 3 Evaluate Using Half-Angle Formulas
Example 4 Verify Identities
Find the value of
if
and
Use the identity
First find the value of
Subtract.
is between
Find the square root of
each side.
Since
Thus,
is in the first quadrant, cosine is positive.
Now find
Double-angle formula
Simplify.
Answer: The value of
Find the value of
if
and
is between
Double-angle formula
Simplify.
Answer: The value of cos 2
Find the value of each expression if
is between
a.
Answer:
b.
Answer:
and
Find
is in the
second quadrant.
Since
we must find
first.
Simplify.
Take the square root
of each side.
Since
is in the second quadrant,
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
Simplify the radicand.
Answer:
Simplify
the denominator.
Find the exact value of
half-angle formulas.
by using the
Simplify the radicand.
Simplify
the denominator.
Answer: Since
is negative. Thus,
is in the third quadrant,
Find the exact value of each expression by using the
half-angle formulas.
a.
Answer:
b.
Answer:
Verify that
an identity.
is
Answer:
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
Original equation
Subtract
Factor.
or
Zero Product Property
Solve.
Look at the graph of
to find solutions of
The solutions are
and so on, and
so on.
The only solution in the interval 0 to
are
and
The period of the sine function is
radians. So the
solutions can be written as
and
where k is any integer.
Similarly, the solutions for
Answer: The solutions are
and
Solve
for all values of
measured in degrees.
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.
if
is
Answer:
b. Solve
measured in degrees.
Answer:
for all values of
if
is
Solve
Original equation
Multiply.
Factor.
or
Check
is undefined.
Thus, is not
a solution.
Answer: The solution is
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 from
each side.
Divide each side
by 6.5.
Add 1.333 to
each side.
Divide each side
by 0.0168.
Answer: Rhonda can plant her flowers at about the 114th
day of the year, or around April 24th.
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 is the day of
the year.
Answer: around the 126th day of the year, or around
May 6th
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