copyright © 2009 pearson addison-wesley 6.3-1 6 inverse circular functions and trigonometric...

Copyright © 2009 Pearson Addison-Wesley 6.3-1

6Inverse Circular Functions and Trigonometric Equations


6.1 Inverse Circular Functions

6.2 Trigonometric Equations I

6.3 Trigonometric Equations II

6.4 Equations Involving Inverse Trigonometric Functions

6Inverse Circular Functions and Trigonometric Equations

Copyright © 2009 Pearson Addison-Wesley

1.1-3

6.3-3

Trigonometric Equations II6.3Equations with Half-Angles ▪ Equations with Multiple Angles


1.1-4

6.3-4

Example 1 SOLVING AN EQUATION USING A HALF-ANGLE IDENTITY

The two numbers over the interval with sine

value

(a) over the interval and

(b) give all solutions.


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6.3-5

Example 1 SOLVING AN EQUATION USING A HALF-ANGLE IDENTITY (continued)

This is a sine curve with period The x-intercepts are the solutions found in Example 1. Using Xscl =makes it possible to support the exact solutions by counting the tick marks from 0 on the graph.


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6.3-6

Example 2 SOLVING AN EQUATION WITH A DOUBLE ANGLE

or

Factor.


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6.3-7

Caution

In the solution of Example 2, cos 2x cannot be changed to cos x by dividing by 2 since 2 is not a factor of cos 2x.

The only way to change cos 2x to a trigonometric function of x is by using one of the identities for cos 2x.


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6.3-8

Example 3 SOLVING AN EQUATION USING A MULTIPLE-ANGLE IDENTITY

Solution set: {30°, 60°, 210°, 240°}

From the given interval 0° ≤ θ < 360°, the interval for 2θ is 0° ≤ 2θ < 720°.


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6.3-9

Example 4 SOLVING AN EQUATION WITH A MULTIPLE ANGLE

Solve tan 3x + sec 3x = 2 over the interval

One way to begin is to express everything in terms of secant.

Square both sides.


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6.3-10

Example 4 SOLVING AN EQUATION WITH A MULTIPLE ANGLE (continued)

Multiply each term of the inequality by 3 to find the interval for 3x:

Using a calculator and the fact that cosine is positive in quadrants I and IV, we have


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6.3-11

Example 4 SOLVING AN EQUATION WITH A MULTIPLE ANGLE (continued)

Since the solution was found by squaring both sides of an equation, we must check that each proposed solution is a solution of the original equation.

Solution set: {.2145, 2.3089, 4.4033}


Frequencies of Piano Keys

A piano string can vibrate at more than one frequency. It produces a complex wave that can be mathema-tically modeled by a sum of several pure tones.

If a piano key with a frequency of f1 is played, then the corresponding string will vibrate not only at f1, but also at 2f1, 3f1, 4f1, …, nf1.

f1 is called the fundamental frequency of the string, and higher frequencies are called the upper harmonics. The human ear will hear the sum of these frequencies as one complex tone.Source: Roederer, J., Introduction to the Physics and Psychophysics of Music, Second Edition, Springer-Verlag, 1975.


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Example 5 ANALYZING PRESSURES OF UPPER HARMONICS

Suppose that the A key above middle C is played on a piano. Its fundamental frequency is f1 = 440 Hz and its associate pressure is expressed as

The string will also vibrate atf2 = 880, f3 = 1320, f4 = 1760, f5 = 2200, … Hz.

The corresponding pressures are


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Example 5 ANALYZING PRESSURES OF UPPER HARMONICS (continued)

The graph of P = P1 + P2 + P3 + P4 + P5 is “saw-toothed.”

(a) What is the maximum value of P?

(b) At what values of t = x does this maximum occur over the interval [0, .01]?


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6.3-15

Example 5 ANALYZING PRESSURES OF UPPER HARMONICS (continued)

A graphing calculator shows that the maximum value of P is approximately .00317.

The maximum occurs at t = x ≈ .000188, .00246, .00474, .00701, and .00928.

copyright © 2009 pearson addison-wesley 6.3-1 6 inverse circular functions and trigonometric...

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