#### Transcript 6.3

```6
Inverse
Circular
Functions
and
Trigonometric
Equations
6.3-1
Inverse Circular Functions
6 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
6.3-2
6.3 Trigonometric Equations II
Equations with Half-Angles ▪ Equations with Multiple Angles
1.1-3
6.3-3
Example 1
SOLVING AN EQUATION USING A
HALF-ANGLE IDENTITY
(a) over the interval
and
(b) give all solutions.
The two numbers over the interval
value
1.1-4
with sine
6.3-4
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.
1.1-5
6.3-5
Example 2
SOLVING AN EQUATION WITH A
DOUBLE ANGLE
Factor.
or
1.1-6
6.3-6
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.
1.1-7
6.3-7
Example 3
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°}
1.1-8
6.3-8
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.
1.1-9
6.3-9
Example 4
SOLVING AN EQUATION WITH A
MULTIPLE ANGLE (continued)
Multiply each term of the inequality
find the interval for 3x:
by 3 to
Using a calculator and the fact that cosine is positive
in quadrants I and IV, we have
1.1-10
6.3-10
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}
1.1-11
6.3-11
Frequencies of Piano Keys
A piano string can vibrate at more than one frequency.
It produces a complex wave that can be mathematically 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.
6.3-12
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
1.1-13
6.3-13
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]?
1.1-14
6.3-14
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.