Copyright © 2017, 2013, 2009 Pearson Education, Inc. 6 Inverse Circular Functions and Trigonometric Equations Copyright © 2017, 2013, 2009 Pearson Education, Inc. 1

Trigonometric Equations II 6.3 Trigonometric Equations II ▪ Equations with Half-Angles ▪ Equations with Multiple Angles ▪ Applications

(a) over the interval and (b) for all solutions. Example 1 SOLVING AN EQUATION WITH A HALF-ANGLE (a) over the interval and (b) for all solutions. The two numbers over the interval with sine value

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(a). Using Xscl = makes it possible to support the exact solutions by counting the tick marks from 0 on the graph.

Example 2 SOLVING AN EQUATION USING A DOUBLE ANGLE IDENTITY Factor. or

Example 3a SOLVING AN EQUATION USING A MULTIPLE-ANGLE IDENTITY From the given interval 0° ≤ θ < 360°, the interval for 2θ is 0° ≤ 2θ < 720°. Solution set: {30°, 60°, 210°, 240°}

Solve for all solutions. Example 3b SOLVING AN EQUATION USING A MULTIPLE-ANGLE IDENTITY Solve for all solutions. All angles 2θ that are solutions of the equation are found by adding integer multiples of 360° to the basic solution angles, 60° and 120°. Solution set, where 180º represents the period of sin2θ: {30° + 180°n, 60° + 180°n, where n is any integer}

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

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

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: {0.2145, 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 mathematically be modeled by a sum of several pure tones. When a piano key with a frequency of f1 is played, the corresponding string vibrates not only at f1 but also at 2f1, 3f1, 4f1, …, nf1. f1 is the fundamental frequency of the string, and higher frequencies are 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.)

The corresponding pressures are 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 at f2 = 880, f3 = 1320, f4 = 1760, f5 = 2200, … Hz. The corresponding pressures are

The graph of P = P1 + P2 + P3 + P4 + P5 is “saw-toothed.” 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, 0.01]?

Example 5 ANALYZING PRESSURES OF UPPER HARMONICS (continued) A graphing calculator shows that the maximum value of P is approximately 0.00317. The maximum occurs at t = x ≈ 0.000191, 0.00246, 0.00474, 0.00701, and 0.00928.