lesson 10.1 defining the circular functions

L E S S O N

10.1CONDENSED

In this lesson, you

● Learn how the circular functions x � cos t and y � sin t are defined

● Find the domain, range, and period of x � cos t and y � sin t

● Find sine and cosine values by using reference angles

Many phenomena—including the ocean tides and the movement of a horse on acarousel—follow repetitive, or cyclical, patterns. You can use the sine and cosinefunctions to model these phenomena.

Investigation: Paddle WheelWork through the investigation in your book, and then compare your results withthose below.

Step 1 One rotation is 360°, so the frog rotates 360° in 6 minutes, or60° per minute. This is 1° per second.

Step 2 You can find the x- and y-values of any point by tracing the graph orby substituting values for t in the equations for x and y.

[�2.35, 2.35, 1, �1.55, 1.55, 1]

Step 3

Here are a few things you can observe in the table: The x- and y-values repeatafter 360°. The x- and y-values are cyclical. For degree values in Quadrant I,

t x y

360 1 0

375 0.966 0.259

390 0.866 0.5

405 0.707 0.707

420 0.5 0.866

435 0.259 0.966

450 0 1

465 �0.259 0.966

480 �0.5 0.866

495 �0.707 0.707

510 �0.866 0.5

t x y

180 �1 0

195 �0.966 �0.259

210 �0.866 �0.5

225 �0.707 �0.707

240 �0.5 �0.866

255 �0.259 �0.966

270 0 �1

285 0.259 �0.966

300 0.5 �0.866

315 0.707 �0.707

330 0.866 �0.5

345 0.966 �0.259

t x y

0 1 0

15 0.966 0.259

30 0.866 0.5

45 0.707 0.707

60 0.5 0.866

75 0.259 0.966

90 0 1

105 �0.259 0.966

120 �0.5 0.866

135 �0.707 0.707

150 �0.866 0.5

165 �0.966 0.259

Defining the Circular Functions

Discovering Advanced Algebra Condensed Lessons CHAPTER 10 155©2004 Key Curriculum Press

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Lesson 10.1 • Defining the Circular Functions (continued)

the x- and y-values are positive. In Quadrant II, the x-values are negative andy-values are positive. In Quadrant III, the x-values are negative and y-values arenegative. In Quadrant IV, the x-values are positive and y-values are negative. Asx-values increase, y-values decrease, and vice versa. The pairing of x- and y-valuesis always the same. That is, �0.866 is always paired with �0.5, and so on.

Step 4

a. The pattern repeats every 360° (or 360 s). Because 1215° � 3(360°) � 135°,the frog’s location at 1215 s is the same as its location at 135 s, which is(�0.707, 0.707). The frog is also at this location at 360 � 135, or 495 s,and at 2(360) � 135, or 855 s.

b. The table shows that the frog is at a height of �0.5 m at 210 s and at330 s. Because of the cyclical pattern, the frog is also at this height at thesetimes plus multiples of 360 s. For the first three rotations, these times are210 s, 330 s, 570 s, 690 s, 930 s, and 1050 s.

c. �1 � y � 1, �1 � x � 1

Step 5 The first graph below shows x � cos t. The second graph shows y � sin t.Both graphs show the same cyclical pattern, although cos t starts at 1 and sin tstarts at 0. They both pass through 1 full cycle and return to the starting locationafter 360°. To find the frog’s position at time t, find the corresponding x-value onthe first graph and y-value on the second graph.

A circle with radius 1 unit centered at the origin is called a unit circle. Using theunit circle in the investigation, you discovered that the values of the sine andcosine functions repeat in a regular pattern. When the output values of a functionrepeat in a regular pattern, the function is periodic. The period of a function isthe smallest distance between values of the independent variable before the cyclebegins to repeat.

t

y

–1

1

90

180

270 360

–1

1

t

x

36027018090

156 CHAPTER 10 Discovering Advanced Algebra Condensed Lessons

©2004 Key Curriculum Press

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Read Example A in your book, which demonstrates that the cosine function has aperiod of 360°. Then read the text between Examples A and B which demonstratesthat the sine function also has a period of 360°. Look closely at the diagram. Makesure you understand the meanings of standard position, terminal side, andreference angle. Read Example B carefully, and then read the example below.

EXAMPLE Find the value of cosine or sine for each angle.

a. cos 225°

b. sin �290°

� Solution For each angle in a and b, rotate the terminal side counterclockwise from thepositive x-axis, then draw a right triangle by drawing a line perpendicular tothe x-axis. Then identify the reference angle.

a. For 225°, the reference angle measures 45°. The x-values in Quadrant IIIare negative, so cos 225° � �cos 45°. Using what you know about theratio of side lengths in a 45°-45°-90° triangle, or using a calculator,cos 225° � � � �0.707.

b. Because the angle measure �290° is negative, rotate the terminal side clockwise290°. The reference angle measures 70°. Because y-coordinates of points inQuadrant I are positive, sin �290° � sin 70° � 0.940.

Angles in standard position are coterminal if they share the same terminal side.For example, the angles measuring 70°, �290°, and 430° are coterminal. Greekletters such as � (theta) and � (alpha) are often used to represent angle measures.

xx

y

y

–290°

70°

xx

y

y1

45°

225°

�2��2

Lesson 10.1 • Defining the Circular Functions (continued)


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L E S S O N

10.2CONDENSED

In this lesson, you

● Calculate arc lengths

● Convert angle measures between degrees and radians

● Find the area of a sector of a circle

● Calculate the angular speed of an object that follows a circular path

So far, you have worked with angles measured in degrees. In this lesson, you willlearn about a different angle measure.

Investigation: A Circle of RadiansRecall from geometry that the measure of an arc is not the same as thelength of an arc. For example, all the arcs marked below have the same measure, but their lengths increase as the circles get bigger.

You can find the length of an arc using the following relationship.

� , or �2�s

r� � �36A0°�

Find the length, s, of each arc pictured in Step 1 in your book. Record your results in a table, and then compare them with those below.(Ignore the values in the last column for now.)

Angles can be measured in a unit called radians.A full circle, or one rotation, is 2� radians, so a halfcircle, or half rotation, is � radians, a quarter rotationis �

�2� radians, and so on.

You can use either of these equivalent relationshipsto convert degrees to radians, or vice versa.

�

��angle in

�radians�

Convert the angles in your table from degrees toradians. Compare your results with those in the� column at right. (Note: Using the conversionrelationships, you can write the formula � � �1

�80� � A,

where � is the angle measure in radians and A is themeasure in degrees.)

angle in degrees��180

angle in radians��2�

angle in degrees��360

A (deg) r (cm) s (cm) � (radians)

90 3 �64��

32��

�2�

90 6 �12

4�� 3� �

�2�

180 2 �42�� 2� �

180 4 �82�� 4� �

45 4 �88��

�4�

45 8 �16

8�� 2� �

�4�

60 4 �86��

43��

�3�

60 6 �12

6�� 2� �

�3�

measure of intercepted angle��360°

length of arc��circumference of circle

r

sA

45°

Radian Measure and Arc Length


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Lesson 10.2 • Radian Measure and Arc Length (continued)

The ratio of the arc length to the circumference is equal to the ratio of theintercepted angle measure to the measure of a complete rotation, regardlessof whether the angle measures are in degrees or radians. So,

�2�s

r� � �2��, or s � r�

(Make sure you do the algebra to verify the equivalence.) So, you can find thelength of an arc by simply multiplying the radius by the angle measure in radians.And because s � r� is equivalent to � � �r

s�, you can find the measure of an

intercepted angle by dividing the arc length by the radius.

Note that it is not necessary to label radian measures with units. To practiceconverting between degrees and radians, complete parts a–d of Example A in yourbook. Then, compare your results with the solutions.

The text on page 576 of your book shows how you can use dimensional analysisto convert between degrees and radians. Read this text carefully. Then, readExample B, which shows you how to find the area of a sector of a circle. Here isanother example.

EXAMPLE Circle P has radius 12 cm, and the measure of central angle DPE is �

23�� radians. What is s, the length of intercepted

arc DE� ? What is the area of the shaded sector?

� Solution To find s, substitute r � 12 cm and � � �23�� radians into

the formula for arc length.

s � r� � 12 � �23�� 8� cm

To find the area of the sector, use the fact that

�AA

s

c

e

i

c

rc

to

le

r� �

The area of the circle is �r2, or 144�. So,

�A14

se

4ct

�or� � � �

13�

Asector � �13� � 144� � 48�

So, the area of the sector is 48� cm2.

Read the remainder of the lesson in your book. Make sure you understand thedefinition of angular speed.

�23��

�2�

measure of intercepted arc��2�

P

12 cm

E

D

s



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L E S S O N

10.3CONDENSED

In this lesson, you

● Find equations for sinusoids

● Identify the amplitude, period, and phase shift of a sinusoid

● Model real data with a sinusoidal function

● Find equations for transformations of the tangent function

Graphs of y � sin x and y � cos x and transformations of these graphs arecalled sine waves or sinusoids. Example A in your book illustrates thattransforming y � sin x is much like transforming any other function.Read that example carefully.

The amplitude of a sinusoid is half the difference of the maximum andminimum function values. This is the same as the absolute value of the scalefactor, or b. The horizontal translation of a sine or cosine graph is called thephase shift. In Example A, the function y � 3 � 2 sin(x � �) has an amplitudeof 2 and a phase shift of ��.

To practice transforming the cosine graph, work through the example below.

EXAMPLE The graph of one cycle (0 � x � 2�) of y � cos x is shown below. Sketch thegraph of one cycle of y � �1 � 3 cos�x � �

�2��. Give the amplitude, period, and

phase shift of the transformed function.

� Solution The coefficient of 3 means that the graph of y � cos x is stretched vertically bya factor of 3. The graph is also translated left �

�2� units and down 1 unit.

The period of the transformed function is 2�, the amplitude is 3, and the phaseshift is ��

�2�.

x

y

2

–4

–2�_2

�

�_2

3�__2

–

x

y

2�

2

–2�_2

3�__2

�

Graphing Trigonometric Functions


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Lesson 10.3 • Graphing Trigonometric Functions (continued)

Investigation: The Pendulum IIRead the investigation in your book. If you have the necessary equipment, collectyour own data and fit a sine function and a cosine function. If not, you can usethese data:

Here is a plot of the data:

[0, 2, 0.1, 0, 1, 0.1]

First, fit a cosine function. The maximum distance value is 0.903, and theminimum is 0.697, so the amplitude of the function is �

12�(0.903 – 0.697), or 0.103.

This is the vertical stretch factor.

A complete cycle from high point to high point goes from 0.10 to 1.475, so theperiod is 1.375. The horizontal scale factor that stretches 2� to 1.375 is �1.

23�75�.

The function y � cos x has a high point at x � 0. This curve has a high point at0.10. So, the phase shift is 0.10.

For y � cos x, the y-value midway between the minimum and maximum valuesis 0. For this curve, it is �

12�(0.903 � 0.697), or 0.8. Thus, the vertical translation

is 0.8.

Putting all this information together gives the function

y � 0.103 cos��1.23�75�(x � 0.1)� � 0.8

Time (s) Distance (m)

1.45 0.894

1.50 0.894

1.55 0.889

1.60 0.879

1.65 0.863

1.70 0.844

1.75 0.824

1.80 0.802

1.85 0.779

1.90 0.760

1.95 0.740

2.00 0.724


0.75 0.698

0.80 0.697

0.85 0.702

0.90 0.712

0.95 0.727

1.00 0.745

1.05 0.766

1.10 0.786

1.15 0.809

1.20 0.830

1.25 0.850

1.30 0.867

1.35 0.881

1.40 0.891


0.05 0.900

0.10 0.903

0.15 0.899

0.20 0.888

0.25 0.873

0.30 0.855

0.35 0.836

0.40 0.814

0.45 0.791

0.50 0.768

0.55 0.749

0.60 0.730

0.65 0.714

0.70 0.704



(continued)

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A transformation of the sine function has the same scale factors and vertical shift,but the phase shift is �

�2� � 0.1, so the sine function is

y � 0.103 sin��1.23�75��x � ��

�2� � 0.1�� 0.8

In these equations, 0.8 represents the average distance from the motion sensorto the washer, 0.103 is the distance from this average to the minimum ormaximum distance, 0.10 is the number of seconds before the washer was first atthe maximum distance, and 1.375 is the number of seconds it takes to completeone cycle.

Example B in your book presents another situation that can be modeled with asinusoidal function. Try to solve the problem posed in that example before youread the solution.

So far in this chapter, you have worked only with sine and cosine. The tangent ofangle A is the ratio of the y-coordinate to the x-coordinate of a point rotated A°(or A radians) counterclockwise from the positive x-axis.

tan A �

Below is the graph of the function y � tan x.

Notice that tan A is undefined for points on the circle with x-coordinate 0. This isshown on the graph by the vertical asymptotes at ��

�2�, �

�2�, �

32��, and so on.

Example C in your book finds the equation for a transformation of y � tan x.Read that example carefully.

x

y

� 2�–�

2

4

–2

–4

6

–6

�_2

3�__2

x-coordinate

Ax

(x, y)

y

y-coordinate

y-coordinate��x-coordinate

Lesson 10.3 • Graphing Trigonometric Functions (continued)


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L E S S O N

10.4CONDENSED

In this lesson, you

● Sketch and examine the graphs of the inverses of y � sin x and y � cos x

● Define the functions y � sin�1 x and y � cos�1 x by restricting the rangesof x � sin y and x � cos y

● Solve equations involving trigonometric functions

The sine, cosine, and tangent functions have repeating values. So, for example, ifyou want to find an angle whose cosine is 0.75, there will be many answers. Forthis reason, using an inverse function on your calculator will not always give youthe angle you are looking for. This is illustrated in Example A in your book. Readthis example carefully.

In previous chapters, you saw that you find the inverse of a relation byexchanging the x- and y-coordinates of all the points. A graph and its inverse arereflections of each other across the line y � x. Page 595 of your book showsgraphs of the exponential function y � bx and its inverse, x � by (or y � logb x),and of the equation y � x2 and its inverse, x � y2. In the case of y � bx, theinverse is a function. In the case of y � x2, it is not. In the investigation, you willexplore the inverses of trigonometric functions.

Investigation: Exploring the InversesComplete the investigation in your book, and then compare your results withthose below.

Steps 1–4 Below are the graphs of y � sin x and x � sin y.

The graph of x � sin y is not a function because there is more than one y-valuefor each x-value. The portion of the graph between y � ��

�2� and y � �

�2� has been

darkened. This portion of the graph is a function because there is exactly oney-value for each x-value.

x

y

� 2�–�–2�

5

–3� 3�

–5

–10

10

Inverses of TrigonometricFunctions


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Lesson 10.4 • Inverses of Trigonometric Functions (continued)

Step 5 Below are the graphs of y � cos x and x � cos y. Any y-interval of theform n� � y � (n � 1)�, where n is an integer, contains a portion of the graphthat is a function. One possibility is 0 � y � �.

The function y � sin�1 x is the portion of the graph of x � sin y from��

�2� � y � �

�2�. (This is the portion you shaded in the investigation.) Similarly,

the function y � cos�1 x is the portion of the graph of x � cos y from 0 � y � �.Restricting the interval guarantees that there is one y-value for every x-value.So, for example, although the equation sin x � 0.5 has infinitely many solutions,the equation x � sin�1(0.5) has only one solution: �

�6�. The value �

�6� is called the

principal value of sin�1(0.5).

Example B in your book shows you how to solve an equation involving atrigonometric function. Work through the example with a pencil and paper.Then test your understanding of the ideas by trying to solve the problem inthe example below.

EXAMPLE Find the first four positive values of x for which 1� 4 sin��x �2

�� 3.

� Solution Graphically, this is equivalent to finding the first four positive intersections of

y � 1 � 4 sin��x �2

�� and y � 3. You can find the approximate intersections shown

below by tracing a graph.

x

y

� 7�2� 3� 4� 5� 6�

2

4

–2

–4

(4.189, 3) (8.378, 3) (16.755, 3) (20.944, 3)

x

y

�

2�

–�

–2�

5

–3� 3�

–5

–10

10



(continued)

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Solving the equation symbolically, you will find one solution.

1 � 4 sin��x �2

�� 3

�4 sin��x �2

�� 2

sin��x �2

��

12�

�x �

2�

� � sin�1��12��

x � 2 sin�1��12��

The unit circle shows that sin�1��12��

�6�. �Remember, the function y � sin�1 x

has a range of ��2� � y � �

�2�.�

So, one solution is x � 2��6��

43��. However, you are looking for positive

solutions. Because the period of y � 1 � 4 sin��x �2

�� is 4�, ��

43�� 4�, or �

83��, is a

solution. This is about 8.378, which corresponds to the second positive solution

marked on the graph.

You can use the symmetry of the graph to find the first positive solution. Thefirst positive solution is the same distance from 2� as the second solution, �

83��.

This distance is �23��. So, 2� � �

23��, or �

43��, is also a solution. This is about 4.189.

Using the fact that the period is 4�, the next two solutions are

x � �43�� 4� � �

163�� 16.755

x � �83�� 4� � �

203�� 20.944

x

1

y

–0.5

– �__6

Lesson 10.4 • Inverses of Trigonometric Functions (continued)


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L E S S O N

10.5CONDENSED

In this lesson, you

● Interpret trigonometric equations that model real situations

● Write trigonometric equations to model real situations

● Find frequencies of periodic functions

Example A in your book shows how the height of water at the mouth of a rivercan be modeled with a trigonometric equation. Work through the examplecarefully. Make sure you understand how the numbers in the equation correspondto the real-life situation.

When the independent variable is time, the period of a function is the length oftime it takes the function to complete one cycle. The frequency of a function isthe reciprocal of the period. It is the number of cycles completed in one unit oftime. For example, if a wave has a period of 0.1 second, then it has a frequency of10 cycles per second.

Investigation: A Bouncing SpringRead the investigation in your book. If you have the necessary equipment, collectthe data and complete the steps of the investigation. If not, complete the stepsusing these sample data. The results given here are based on the sample data.

Time Height(s) (m)

2.52672 0.609038

2.58048 0.632104

2.63424 0.657778

2.688 0.682628

2.74176 0.702261

2.79552 0.713794

2.84928 0.714206

2.90304 0.705419

2.9568 0.690454

3.01056 0.667938

3.06432 0.64144

3.11808 0.618374

3.17184 0.60588

3.2256 0.597505

3.27936 0.597917

Time Height(s) (m)

1.72032 0.607665

1.77408 0.630044

1.82784 0.665192

1.8816 0.699241

1.93536 0.720659

1.98912 0.731917

2.04288 0.728622

2.09664 0.715167

2.1504 0.687845

2.20416 0.664505

2.25792 0.631417

2.31168 0.607391

2.36544 0.592975

2.4192 0.586934

2.47296 0.592837

Time Height(s) (m)

0.86016 0.587071

0.91392 0.597093

0.96768 0.640204

1.02144 0.668624

1.0752 0.705831

1.12896 0.729308

1.18272 0.741253

1.23648 0.738919

1.29024 0.724366

1.344 0.695946

1.39776 0.656954

1.45152 0.620983

1.50528 0.587895

1.55904 0.567712

1.6128 0.5673

1.66656 0.581991

Time Height(s) (m)

0 0.561397

0.05376 0.563182

0.10752 0.578284

0.16128 0.604919

0.21504 0.639792

0.2688 0.673841

0.32256 0.699653

0.37632 0.71805

0.43008 0.724229

0.48384 0.719698

0.5376 0.729446

0.59136 0.701163

0.64512 0.632653

0.69888 0.603409

0.75264 0.576362

0.8064 0.564005

Modeling with TrigonometricEquations


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Lesson 10.5 • Modeling with Trigonometric Equations (continued)

Step 2 At right is a graph of the data.

The average maximum is 0.728, and the average minimum is 0.575, sothe amplitude is �0.728 �

20.575� � 0.077. The average value, which is the

vertical translation, is 0.652. The periods of the four cycles are 0.808,0.805, 0.806, and 0.860, so 0.807 may be a good choice. The frequencyis then �0.8

107�, or 1.239 cycles per second. The first maximum occurs at

t � 0.43, so if you choose a cosine function, it should have a phase [0, 3.3, 1, �0.1, 0.9, 1]

shift of 0.43.

Step 3 h � 0.652 � 0.077 cos��2�(t0.

�80

07.43)

��A graph of the curve with the data shows a good fit.

[0, 3.3, 1, �0.1, 0.9, 1]

Step 4

a. 0.652 m is the average height of the spring. 0.077 m is the amount up anddown from the average height the spring moves. 0.807 s is the length oftime it takes to complete one cycle. 0.43 s is the time when the firstmaximum occurs.

b. If you moved the sensor 1 m farther away, the vertical translation wouldincrease by 1. All other values would remain the same.

c. If you pulled the spring down lower, the amplitude would increase. Theperiod might change too.

Example B in your book presents another situation that can be modeled witha periodic function. Work through the example using a pencil and paper. Theexample below is Exercise 8a in your book. Try to write the equation withoutlooking at the solution.

EXAMPLE The time between high and low tide in a river harbor is approximately 7 h. Thehigh-tide depth of 16 ft occurs at noon and the average harbor depth is 11 ft.Write an equation modeling this relationship.

� Solution The tide completes half a cycle in 7 h, so the period is 14 h. The horizontalscale factor that stretches 2� to 14 is �

�7

�. The average depth, 11, is the verticaltranslation. The amplitude is 5, the difference between the high-tide depthand the average depth. If you assume that t � 0 corresponds to noon, thenthe function starts at a high point. So, if you use the cosine function, thereis no phase shift. The equation is

d � 11 � 5 cos ��7t

�



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L E S S O N

10.6CONDENSED

In this lesson, you

● Define the cotangent, secant, and cosecant functions

● Apply the reciprocal identities and use them to prove other trigonometricidentities

● Derive and prove three Pythagorean identities

An identity is an equation that is true for all the values for which the expressions

are defined. For example, sin A � cos�A � ��2�� is an identity because it is true no

matter what value you substitute for A. Read the lesson introduction in your

book, which shows that tan A � �csions

AA� is an identity.

The reciprocals of the tangent, cosine, and sine functions are also trigonometricfunctions. The reciprocal of the tangent is the cotangent, abbreviated cot. Thereciprocal of the cosine is the secant, abbreviated sec. The reciprocal of the sineis the cosecant, abbreviated csc. These definitions lead to the six reciprocalidentities below.

cot A � �tan1

A� sec A � �co1s A� csc A � �sin

1A�

tan A � �co1t A� cos A � �se

1c A� sin A � �cs

1c A�

Your calculator does not have special keys for the cotangent, secant, andcotangent. So, for example, to graph y � cot x, you must use y � �ta

1n x�. To

familiarize yourself with the new functions, graph each pair of reciprocalfunctions, one pair at a time, on your calculator (that is, graph cotangent andtangent, then secant and cosine, then cosecant and sine).

One method of proving an identity involves writing equivalent expressions forone side of the equation until it is the same as the other side. You can use anyidentities you have already proved. This is demonstrated in the example in yourbook. Work through that example with a pencil and paper.

Investigation: Pythagorean IdentitiesTry completing the investigation on your own. The answers are given below if youneed them.

Step 1 The graph of y � sin2 x � cos2 x is the horizontal line y � 1. Based onthis graph, you can write the identity sin2 x � cos2 x � 1.

[0, 2�, ��2, �3, 3, 1]

Step 2 The lengths of the legs of the triangle shown are sin A and cos A, and thehypotenuse has length 1. By the Pythagorean Theorem, sin2 A � cos2 A � 1.

Fundamental TrigonometricIdentities


(continued)

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Lesson 10.6 • Fundamental Trigonometric Identities (continued)

Step 3 The equation sin2 x � cos2 x � 1 is called a Pythagorean identity becauseit is derived by using the Pythagorean Theorem. (In a unit circle with a referencetriangle, sin x and cos x are the lengths of the legs of the right triangle and 1is the length of the hypotenuse. So, using the Pythagorean Theorem,sin2 x � cos2 x � 1.)

Step 4

cos2 x � 1 � sin2 x

sin2 x � 1 � cos2 x

Step 5 The identity is tan2 x � 1 � sec2 x. The derivation below serves as aproof, but you should practice the method of the example by manipulating theleft side of the identity, tan2 x � 1, until it equals the right side, sec2 x.

sin2 x � cos2 x � 1 Original identity.

� � Divide both sides by cos2 x.

� � sin2 x means (sin x)2, and cos2 x means (cos x)2.

� �2

� � �2

� � �2

�ab 2

2� � ��

ab��

2

tan2 x � 1 � sec2 x Use identities �csions

xx� � tan x and �co

1s x� � sec x.

Step 6 Both y � tan2 x � 1 and y � sec2 x have the same graph, which verifies the identity. The identity is undefined when cos x � 0, or whenx � �

�2� � �n or x � 90° � 180°n, where n is an integer.

Step 7 The identity is 1 � cot2 x � csc2 x. The derivation below serves as a proof:

sin2 x � cos2 x � 1 Original identity. [0, 2�, ��2, �3, 3, 1]

� � Divide both sides by sin2 x.

� � sin2 x means (sin x)2, and cos2 x means (cos x)2.

� �2

� � �2

� � �2

�ab 2

2� � ��

ab��

2

1 � cot2 x � csc2 x Use identities �csions

xx

� � cot x and �sin1

x� � csc x.

Step 8 Both y � 1 � cot2 x and y � csc2 x have the same graph, whichverifies the identity. The identity is undefined when sin x � 0, or when x � �n or x � 180°n, where n is an integer.

[0, 2�, ��2, �3, 3, 1]

The Pythagorean identities you proved in the investigation are summarized onpage 611 of your book.

1�sin x

cos x�sin x

sin x�sin x

1�(sin x)2

(cos x)2

�(sin x)2(sin x)2

�(sin x)2

1�sin2 x

cos2 x�sin2 x

sin2 x�sin2 x

1�cos x

cos x�cos x

sin x�cos x

1�(cos x)2

(cos x)2

�(cos x)2(sin x)2

�(cos x)2

1�cos2 x

cos2 x�cos2 x

sin2 x�cos2 x



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L E S S O N

10.7CONDENSED

In this lesson, you

● Model a sound with a sum of two sinusoidal equations

● Develop or prove several trigonometric identities

● Use trigonometric identities to find sines and cosines of angles

The sound of a note played by a musical instrument can be modeled bycombining more than one trigonometric function. Read the text before theinvestigation in your book to find out more.

Investigation: Sound WaveThe investigation asks you to ring a tuning fork and then collect data using amicrophone probe. Below are equations and graphs for data collected from G-392 Hz and C-256 Hz tuning forks.

G: y � 0.116 sin��0.020�25�(x � 0.0027)� � 2.681

[0, 0.02, 1, 2.2, 3, 1]

C: y � 0.0828 sin��0.020�38�(x � 0.0012)� � 2.6815

[0, 0.02, 1, 2.2, 3, 1]

Simply adding the equations of the single C and G notes gives a close approximation of the wave formed by playing the two notes together,except that the curve is shifted up because the two horizontal shiftsare combined.

Subtracting 2.6815 from the sum of the equations will make the newequation match the horizontal shift of the new data.

y � 0.116 sin��0.020�25�(x � 0.0027)� � 2.681 � 0.0828 sin��0.0

20�38�(x � 0.0012)�

Combining TrigonometricFunctions


(continued)

[0, 0.02, 1, 2.2, 6, 1]

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Lesson 10.7 • Combining Trigonometric Functions (continued)

A horizontally translated sinusoid can be written as the sum of twountranslated curves. For example, you can use your calculator toverify that y � cos(x � 0.6435) is equivalent to y � 0.8 cos x � 0.6 sin x.The text on page 617 of your book proves the identity

cos(A � B) � cos A cos B � sin A sin B

Follow along with the steps using a pencil and paper.

Example A in your book shows how you can use the identity above to find exactcosine values for some new angles, using values you already know. Here is anotherexample.

EXAMPLE A Find the exact value of cos �71�2�.

� Solution cos �71�2� � cos��

1102��

31�2�� Rewrite �

71�2� as the difference of two fractions.

� cos��56��

�4�� Reduce.

� cos �56�� cos �

�4� � sin �

56�� sin �

�4� cos(A � B) � cos A cos B � sin A sin B.

� � � � �12� �

Substitute exact values for sine and cosine of �56��

and ��4�.

� Combine terms.

In Example B in your book, the identity cos(A � B) � cos A cos B � sin A sin B isused to develop the identity

cos(A � B) � cos A cos B � sin A sin B

Read that example carefully. The example below develops the identity

sin(A � B) � sin A cos B � cos A sin B

EXAMPLE B Use the identity for cos(A � B) and the identities sin A � cos��2� � A� and

cos A � sin��2� � A� to develop an identity for sin(A � B).

� Solution sin(A � B) � cos��2� � (A � B)� sin A � cos��

�2� � A�.

� cos��2� � A� � B� Rewrite �

�2� � (A � B) as ��

�2� � A� � B.

� cos��2� � A� cos B � sin��

�2� � A� sin B Use the identity for cos(A � B).

� sin A cos B � cos A sin B cos A � sin��2� � A� and

sin A � cos��2� � A�.

More identities are given in the box on page 619 of your book. You will provethese identities in the exercises.

��6� � �2��4

�2��2

�2��2

�3��2



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lesson 10.1 defining the circular functions

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