Copyright © 2009 Pearson Addison-Wesley Inverse Circular Functions and Trigonometric Equations

Copyright © 2009 Pearson Addison-Wesley Inverse Circular Functions 6.2 Trigonometric Equations I 6.3 Trigonometric Equations II 6.4 Equations Involving Inverse Trigonometric Functions 6 Inverse Circular Functions and Trigonometric Equations

Copyright © 2009 Pearson Addison-Wesley Trigonometric Equations II 6.3 Equations with Half-Angles ▪ Equations with Multiple Angles

Copyright © 2009 Pearson Addison-Wesley 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.

Copyright © 2009 Pearson Addison-Wesley 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.

Copyright © 2009 Pearson Addison-Wesley Example 2 SOLVING AN EQUATION WITH A DOUBLE ANGLE or Factor.

Copyright © 2009 Pearson Addison-Wesley 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.

Copyright © 2009 Pearson Addison-Wesley 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°.

Copyright © 2009 Pearson Addison-Wesley 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.

Copyright © 2009 Pearson Addison-Wesley 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

Copyright © 2009 Pearson Addison-Wesley 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, , }

Copyright © 2009 Pearson Addison-Wesley 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 f 1 is played, then the corresponding string will vibrate not only at f 1, but also at 2f 1, 3f 1, 4f 1, …, nf 1. f 1 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.

Copyright © 2009 Pearson Addison-Wesley Example 5 ANALYZING PRESSURES OF UPPER HARMONICS Suppose that the A key above middle C is played on a piano. Its fundamental frequency is f 1 = 440 Hz and its associate pressure is expressed as The string will also vibrate at f 2 = 880, f 3 = 1320, f 4 = 1760, f 5 = 2200, … Hz. The corresponding pressures are

Copyright © 2009 Pearson Addison-Wesley Example 5 ANALYZING PRESSURES OF UPPER HARMONICS (continued) The graph of P = P 1 + P 2 + P 3 + P 4 + P 5 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]?

Copyright © 2009 Pearson Addison-Wesley Example 5 ANALYZING PRESSURES OF UPPER HARMONICS (continued) A graphing calculator shows that the maximum value of P is approximately The maximum occurs at t = x ≈ ,.00246,.00474,.00701, and