lesson 13.1 defining the circular functions - high school...

L E S S O N

13.1CONDENSED

Discovering Advanced Algebra Condensed Lessons CHAPTER 13 191©2010 Key Curriculum Press

In this lesson you will

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

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

● find sine and cosine values by using reference angles

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

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

Step 1 One rotation is 360°, so the frog rotates 360° in 6 minutes, or 60° per minute. This is 1° per second.

Step 2 Make sure your calculator is set to degrees.

Step 3 You can find the x- and y-values of any point by tracing the graph or by substituting values for t in the equations for x and y. The coordinates for the point that the frog reaches at time t are (cos t, sin t). More appropriate names for hpos and vpos might be cosine and sine.

Step 4 The x- and y-values repeat after 360°. The x- and y-values are cyclical. For degree values in Quadrant I, the x- and y-values are positive. In Quadrant II, the x-values are negative and y-values are positive. In Quadrant III, the x-values are negative and y-values are negative. In Quadrant IV, the x-values are positive and y-values are negative. The pairing of x- and y-values is always the same. That is, �0.866 is always paired with �0.5, and so on.

Step 5

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 at 330 s. Because of the cyclical pattern, the frog is also at this height at these times plus multiples of 360 s. For the first three rotations, these times are 210 s, 330 s, 570 s, 690 s, 930 s, and 1050 s.

c. �1 � y � 1, �1 � x � 1

Defining the Circular Functions

(continued)

DAA2CL_010_13.indd 191DAA2CL_010_13.indd 191 1/13/09 5:43:56 PM1/13/09 5:43:56 PM

192 CHAPTER 13 Discovering Advanced Algebra Condensed Lessons

©2010 Kendall Hunt Publishing

Step 6 The graph below shows scatter plots of (time, hpos), and (time, vpos). At 0 seconds the frog is at (1, 0), so the hpos graph starts at (0, 1), whereas the vpos graph starts at (0, 0). Both graphs pass through one full cycle and return to the same starting location after 360 seconds, or 360° around the circle. The graphs appear to be translations of one another. To find the frog’s position at time t, subtract multiples of 360° from t until the result is between 0° and 360°, and then find the y-value on each graph. The frog’s position is (hpos, vpos), or (cos t, sin t). The cosine and sine functions are sometimes called circular functions because their values repeat with every rotation around a circle.

As you’ve learned in earlier lessons, a circle with radius 1 unit centered at the origin is called a unit circle. Using the unit circle in the investigation, you discovered that the values of the sine and cosine functions repeat in a regular pattern. When the output values of a function repeat in a regular pattern, the function is periodic. The period of a function is the smallest distance between values of the independent variable before the cycle begins to repeat.

Read Example A in your book, which demonstrates that the cosine function has a period of 360°. Then read the text between Examples A and B which demonstrates that the sine function also has a period of 360°. Look closely at the diagram on page 738. Make sure you understand the meanings of standard position, terminal side, and reference 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 parts a and b, rotate the point (1, 0) counterclockwise about the origin. Draw a ray from the origin through the image point to be the terminal side. Drop a perpendicular line from the image point to the x-axis to create a reference triangle, and then identify the reference angle.

Lesson 13.1 • Defining the Circular Functions (continued)

(continued)


Discovering Advanced Algebra Condensed Lessons CHAPTER 13 193©2010 Kendall Hunt Publishing

a. For 225°, the reference angle measures 45°. The x-values in Quadrant III are negative, so cos 225° � �cos 45°. Using what you know about the ratio of side lengths in a 45°-45°-90° and the fact that the hypotenuse of the triangle is 1, you can find that the lengths of the legs are both 1

___ �

__ 2 , or �

__ 2 ___ 2 . So the

coordinates of the point are � � � __

2 ____ 2 , � �

__ 2 ____ 2 � and the cosine of the angle is

� � __

2 ____ 2 � �0.707.

x x

y

y1

45°

225°

b. Because the angle measure �290° is negative, rotate the point clockwise 290°. The reference angle measures 70°. Because y-coordinates of points in Quadrant I are positive, sin �290° � sin 70° � 0.940.

x x

y

y

–290°

70°

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

Lesson 13.1 • Defining the Circular Functions (continued)


CONDENSED



● 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 will learn about a different angle measure, called a radian. Before you work on the investigation, you might want to review the formulas for arc measure and arc length by working through the lesson Refreshing Your Skills for Chapter 13 in your book.

Investigation: A Radian ProtractorUse the semicircle at right to complete Steps 1–4 on your own. Then work through the rest of the investigation and compare your results to those below for Steps 5–7.

Step 5 If the radius of the circle is r, then the circumference is 2�r units. So half of the circle has length �r units. This is � radians.

Step 6 Because a right angle intercepts a fourth of the circle, the length of the intercepted arc would be half of the length of a semicircle, or � __ 2 r . This is � __ 2 radians.

Step 7 Using the same reasoning as in Steps 6 and 7, you can find the radian values for the common angles.

30° = � __ 6 radians 45° = � __ 4 radians 60° = � __ 3 radians

120° = 2� ___ 3 radians 135° = 3� ___ 4 radians 150° = 5� ___ 6 radians

Make sure you have everything marked on your radian protractor. Counterclockwise,

you should have marks labeled 0, 0.5, � __ 6 , � __ 4 , 1, � __ 3 , 1.5, � __ 2 , 2, 2� __ 3 , 3�

__ 4 , 2.5, 5� __ 6 , 3, and �.

��6

��4

�

��3

��22��

33��4

5��6

3

2.5

2 1.51

0.5

0

Radian MeasureL E S S O N

13.2

(continued)




In the investigation, you found that a full circle, or one rotation, is 2� radians, so a half circle, or half rotation, is � radians, a quarter rotation is �

�2� radians, and so on.

You can write proportions to express these equivalent relationships.

angle in degrees

_____________ 360 � angle in radians

_____________ 2�

angle in degrees _____________ 180 �

angle in radians _____________ �

Solving either of these proportions for the angle in degrees or the angle in radians will give you a formula you can use to convert degrees to radians, or vice versa.

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

The text on page 746 of your book shows how you can use dimensional analysis to convert between degrees and radians. Read this text carefully.

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

s ____ 2�r � � ___ 2� , or s � r �

You can find the length of an arc by multiplying the radius by the angle measure in radians. And because s � r � is equivalent to � � s _ r , you can find the measure of an intercepted angle by dividing the arc length by the radius.

Now, read Example B, which shows you how to find the area of a sector of a circle. Here is another example.

EXAMPLE Circle P has radius 12 cm, and the measure of central

P

12 cm

E

D

s

angle DPE is 2� __ 3 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 � � 2� __ 3 radians into

the formula for arc length.

s � r � � 12 � 2� ___ 3 � 8� cm

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

A sector _____ A circle

� measure of intercepted arc

_____________________ 2�

The area of the circle is �r 2, or 144�. So,

A sector _____ 144� �

2� ___ 3 ____

2� � 1 __ 3

A sector � 1 __ 3 � 144� � 48�

The area of the sector is 48� cm2.

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

Lesson 13.2 • Radian Measure (continued)


CONDENSED



● 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 are called sine waves or sinusoids. Example A in your book illustrates that transforming 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 and minimum function values. This is the same as the absolute value of the vertical scale factor, or ⏐b⏐. The horizontal translation of a sine or cosine graph is called the phase shift. In Example A, the function y � 3 � 2 sin(x � �) has an amplitude of 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 the graph of one cycle of y � �1 � 3 cos�x � � __ 2 �. Give the amplitude, period, and phase shift of the transformed function.

x

y

2�

2

–2�_2

3�__2

�

� Solution The coefficient of 3 means that the graph of y � cos x is dilated vertically by a factor of 3. The graph is also translated left � __ 2 units and down 1 unit. Pick a few points on the original graph to see how they are transformed. The x-intercepts, � � __ 2 , 0 � and � 3�

__ 2 , 0 � , are not affected by the vertical dilation, because 3 � 0 � 0. Now perform the translation. The image of � � __ 2 , 0 � after a translation left � __ 2 units and

down 1 unit is (0, �1) and the image of � 3� __ 2 , 0 � is (�, �1). Now consider a point

at a low point on the original graph, say (�, �1). The vertical dilation transforms this point into (�, �3). Apply the translation to get � � __ 2 , �4 � . This point is on the transformed function below.

x

y

2

–4

–2�_2

�

�_2

3�__2

–

The period of the transformed function is 2�, the amplitude is 3, and the phase shift is � � __ 2 .

L E S S O N

13.3 Graphing Trigonometric Functions

(continued)




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

Time (s) Distance (m)

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


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


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

Here is a plot of the data:

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

12�(0.903 – 0.697), or 0.103.

This is the vertical scale factor.

A complete cycle from high point to high point goes from 0.10 to 1.475, so the period is 1.375. The horizontal scale factor that dilates 2� to 1.375 is 1.375

____ 2� .

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

For y � cos x, the y-value midway between the minimum and maximum values is 0. For this curve, it is 1 _ 2 (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 � 2� _____ 1.375 (x � 0.1)� � 0.8

Lesson 13.3 • Graphing Trigonometric Functions (continued)

(continued)



A transformation of the sine function has the same scale factors and vertical shift, but the function must shift another quarter cycle, or 1.375

____ 4 . Using the phase shift � � 1.375

____ 4 � 0.1 � , the sine function is

y � 0.103 sin � 2� _____ 1.375 �x � � � 1.375 _____ 4 � 0.1 � � � � 0.8

In these equations, 0.8 represents the average distance from the motion sensor to the washer, 0.103 is the distance from this average to the minimum or maximum distance, 0.1 is the number of seconds before the washer was first at the maximum distance, and 1.375 is the number of seconds it takes to complete one cycle. You can check the fit with your calculator. Make sure your calculator is set to radians.

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

So far in this chapter, you have worked primarily with sine and cosine. The tangent of angle 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 � y-coordinate

__________ x-coordinate

x-coordinate

Ax

(x, y)

y

y-coordinate

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

x

y

� 2�–�

2

4

–2

–4

6

–6

�_2

3�__2

Notice that tan A is undefined for points on the circle with x-coordinate 0. This is shown on the graph by the vertical asymptotes at � � __ 2 , � __ 2 , 3�

__ 2 , and so on.

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

Lesson 13.3 • Graphing Trigonometric Functions (continued)


CONDENSED



● 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 ranges of x � sin y and x � cos y

● solve equations involving trigonometric functions

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

In previous chapters, you saw that you find the inverse of a relation by exchanging the x- and y-coordinates of all the points. A graph and its inverse are reflections of each other across the line y � x. Page 769 of your book shows graphs of the exponential function y � b x and its inverse, x � b y (or y � log b x), and of the equation y � x 2 and its inverse, x � y 2. In the case of y � b x, the inverse is a function. In the case of y � x 2, it is not. In the investigation, you will explore the inverses of trigonometric functions.

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

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

x

y

� 2�

5

5

3�

–5

–10

10

–2�

–5

–3�

–3�

–10

–�

–�

10–2�

�

2�

3�

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

Inverses of Trigonometric Functions

L E S S O N

13.4

(continued)




(continued)

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

x

y

� 2�

5

5

3�

–5

–10

10

–2�

–5

–3�

–3�

–10

–�

–�

10–2�

�

2�

3�

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 a trigonometric function. Work through the example with a pencil and paper. Then test your understanding of the ideas by trying to solve the problem in the 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)

Lesson 13.4 • Inverses of Trigonometric Functions (continued)



Solving the equation symbolically, you will find one solution.

1 � 4 sin � x � � _____ 2 � � 3

�4 sin � x � � _____ 2 � � 2

sin � x � � _____ 2 � � � 1 __ 2

x � � _____ 2 � sin�1 � � 1 __ 2 �

x � 2 sin�1 � � 1 __ 2 � � �

The unit circle shows that sin�1 � � 1 _ 2 � � � � __ 6 . �Remember, the function y � sin�1 x

has a range of � � __ 2 � y � � __ 2 .�

x

1

y

–0.5

– �__6

So, one solution is x � 2 � � � __ 6 � � � � � 4� __ 3 . However, you are looking for positive

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

__ 3 � 4�, or 8� __ 3 , 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. The first positive solution is the same distance from 2� as the second solution, 8�

__ 3 . This distance is 2�

__ 3 . So, 2� � 2� __ 3 , or 4�

__ 3 , is also a solution. This is about 4.189. Using the fact that the period is 4�, the next two solutions are

x � 4� ___ 3 � 4� � 16� ____ 3 � 16.755

x � 8� ___ 3 � 4� � 20� ____ 3 � 20.944

Lesson 13.4 • Inverses of Trigonometric Functions (continued)


CONDENSED



● 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 river can be modeled with a trigonometric equation. Work through the example carefully. Make sure you understand how the numbers in the equation correspond to the real-life situation.

When the independent variable is time, the period of a function is the length of time it takes the function to complete one cycle. The frequency of a function is the reciprocal of the period. It is the number of cycles completed in one unit of time. For example, if a wave has a period of 0.1 second, then it has a frequency of 10 cycles per second.

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

Time (s)

Height (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

Time (s)

Height (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 (s)

Height (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 (s)

Height (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

Modeling with Trigonometric Equations

L E S S O N

13.5

(continued)




Step 2 At right is a graph of the data.

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

_________ 2 � 0.076. The average value, which is the vertical translation, is 0.652. The minima occur at 0, 0.80640, 1.61280, 2.41920, and 3.22560, at intervals of 0.8064, so the period is

0.8064 s. The frequency is then 1 _____ 0.8064 , or 1.240 cycles per second. The

first maximum occurs at t � 0.43, so if you choose a cosine function, it should have a phase shift of 0.43.

Step 3 h � 0.652 � 0.076 cos � 2�(t � 0.43) ________ 0.8064 �

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

Step 4

a. 0.652 m is the average height of the spring. 0.076 m is the amount up and down from the average height the spring moves. 0.8064 s is the length of time it takes to complete one cycle. 0.43 s is the time when the first maximum occurs.

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

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

Example B in your book presents another situation that can be modeled with a periodic function. Work through the example using a pencil and paper. The example below is Exercise 8a in your book. Try to write the equation without looking at the solution.

EXAMPLE The time between high and low tide in a river harbor is approximately 7 h. The high-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 horizontal scale factor that dilates 2� to 14 is 7 __ � . The average depth, 11, is the vertical translation. The amplitude is 5, the difference between the high-tide depth and the average depth. If you assume that t � 0 corresponds to noon, then the function starts at a high point. So if you use the cosine function, there is no phase shift. The equation is

d � 11 � 5 cos �t ___ 7

Lesson 13.5 • Modeling with Trigonometric Equations (continued)


CONDENSED



● define the cotangent, secant, and cosecant functions

● apply the reciprocal identities and use them to prove other trigonometric identities

● 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 � sin A ____ cos A is an identity.

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

cot A � 1 _____ tan A

sec A � 1 _____ cos A

csc A � 1 _____ sin A

tan A � 1 _____ cot A

cos A � 1 _____ sec A

sin A � 1 _____ csc A

Your calculator may not have special keys for the cotangent, secant, and cotangent. For example, to graph y � cot x, you might need to use y � 1

____ tan x .To familiarize yourself with the new functions, graph each pair of reciprocal functions, one pair at a time, on your calculator (that is, graph cotangent and tangent, then secant and cosine, then cosecant and sine).

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

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

Step 1 The graph of y � sin2 x � cos2

x is the horizontal line y � 1. Based on this graph, you can write the identity sin2

x � cos2 x � 1.

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

A � cos2 A � 1.

L E S S O N

13.6Fundamental Trigonometric Identities

(continued)




Step 3 The equation sin2 x � cos2

x � 1 is called a Pythagorean identity because it is derived by using the Pythagorean Theorem. (In a unit circle with a reference triangle, sin x and cos x are the lengths of the legs of the right triangle and 1 is 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 a proof, but you should practice the method of the example by manipulating the left side of the identity, tan2

x � 1, until it equals the right side, sec2 x.

sin2 x � cos2

x � 1 Original identity.

sin2 x _____

cos2 x

� cos2 x _____

cos2 x

� 1 _____ cos2 x

Divide both sides by cos2 x.

(sin x)2

______ (cos x)2 �

(cos x)2 ______

(cos x)2 � 1 ______ (cos x)2 sin2 x means (sin x)2, and cos2 x means (cos x)2.

� sin x ____ cos x � 2 � � cos x ____ cos x � 2 � � 1 ____ cos x � 2 a 2 __

b 2 � � a _

b � 2

tan2 x � 1 � sec2

x Use identities sin x ____ cos x � tan x and 1

____ cos x � sec x.

Step 6 Both y � tan2 x � 1 and y � sec 2 x have the same graph,

which verifies the identity. The identity is undefined when cos x � 0, or when x � � __ 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.

sin2 x _____

sin2 x

� cos2 x _____

sin2 x

� 1 _____ sin2 x

Divide both sides by sin2 x .

(sin x)2

______ (sin x)2 �

(cos x)2 ______

(sin x)2 � 1 ______ (sin x)2 sin2 x means (sin x)2, and cos2 x means (cos x)2.

� sin x ____ sin x

� 2 � � cos x ____ sin x

� 2 � � 1 ____ sin x

� 2 a 2 __

b 2 � � a _

b � 2

1 � cot2 x � csc2

x Use identities cos x ____ sin x � cot x and 1

___ sin x � csc x .

Step 8 Both y � 1 � cot2 x and y � csc2

x have the same graph, which verifies the identity. The identity is undefined when sin x � 0, or when x � �n or x � 180°n, where n is an integer.

The Pythagorean identities you proved in the investigation are summarized on page 785 of your book.

Lesson 13.6 • Fundamental Trigonometric Identities (continued)


CONDENSED


Combining Trigonometric Functions


● 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 by combining more than one trigonometric function. Read the text before the investigation in your book to find out more.

Investigation: Sound WaveThe investigation asks you to ring a tuning fork and then collect data using a microphone probe. Below are equations and graphs for data collected from A-440 Hz and E-329.6 Hz tuning forks.

A: y � 0.01587 sin � 2� _______ 0.00229 (x � 0.0001)� � 0.00696

E: y � 0.02326 sin � 2� _______ 0.00304 (x � 0.0001)� � 0.0062

To write a model for the combined graph, you can add the two sine equations and then adjust their coefficients. The frequency for each sine graph should be approximately the same in the combined equation. However, you’ll need to experiment to find the other coefficients. The following equation fits the data reasonably well.

y � 0.004 sin � 2� _______ 0.00229 � (x � 0.0074)� � 0.0075 � 0.0033 sin � 2� _______ 0.00304

� (x � 0.00687)� � 0.00034

L E S S O N

13.7

(continued)




A horizontally translated sinusoid can be written as the sum of two untranslated curves. For example, you can use your calculator to verify that y � cos(x � 0.6435) is equivalent to y � 0.8 cos x � 0.6 sin x. The text on page 791 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 exact cosine values for some new angles, using values you already know. Here is another example.

EXAMPLE A Find the exact value of cos 7� __ 12 .

� Solution cos 7� ___ 12 � cos � 10� ____ 12 � 3� ___ 12 � Rewrite 7�

__ 12 as the difference of two fractions.

� cos � 5� ___ 6 � � __ 4 � Reduce.

� cos 5� ___ 6 � cos � __ 4 � sin 5� ___ 6 � sin � __ 4 cos(A � B) � cos A cos B � sin A sin B.

� � � __

3 ____ 2 � � __

2 ____ 2 � 1 __ 2 � � __

2 ____ 2 Substitute exact values for sine and cosine of

5�

__

6

and

�

_

4 .

� � � __

6 � � __

2 ___________ 4 Combine terms.

In Example B in your book, the identity cos(A � B) � cos A cos B � sin A sin B is used 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 793 of your book. You will prove these identities in the exercises.

Lesson 13.7 • Combining Trigonometric Functions (continued)


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Please list any comments you have about this book.

Do you have any suggestions for improving the student or teacher material?

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lesson 13.1 defining the circular functions - high school...

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