
698 CHAPTER 6 Trigonometric Identities, Inverses, and Equations 6-84
STRENGTHENING CORE SKILLS
tangent to (bounces off
of) the x-axis. This indi-
cates a total of four beats.
Note the number of beats
is equal to the difference 
Further experimentation
will show this is not a
coincidence, and this
enables us to con-
struct two additional
functions that will
frame these pulsa-
tions and make them
easier to see. Since
the maximum ampli-
tude of the resulting
wave is 2, we use
functions of the
form  to construct the frame, where k is the
 2 cosa
k
2
xb
m  n: 12  8  4.
number of beats in the interval  For
and we have 
and the functions we use will be
and  as shown in
Figure 6.35. The result is shown in Figure 6.36, where
the frame clearly shows the four beats or more precisely,
the four moments of silence.
For each exercise, (a) express the sum  as a
product, (b) graph Y
R
on a graphing calculator for
and identify the number of beats in this inter-
val, and (c) determine what value of k in 
would be used to frame the resultant Y
R
, then enter these
as Y
2
and Y
3
to check the result.
Exercise 1:
Exercise 2:
Exercise 3:
Exercise 4:
Y
1
 cos111t2;
  
Y
2
 cos110t2
Y
1
 cos114t2;
  
Y
2
 cos16t2
Y
1
 cos112t2;
  
Y
2
 cos19t2
Y
1
 cos114t2;
  
Y
2
 cos18t2
2 cosa
k
2
xb
x  30, 24
Y
1
 Y
2
Y
3
2 cos12x2Y
2
 2 cos12x2
k 
12  8
2
 2
Y
2
 cos18t2,Y
1
 cos112t2
1m  n  k2.
Figure 6.35
Figure 6.36
3
3
2
0
College Algebra & Trignometry—
Trigonometric Equations and Inequalities
The ability to draw a quick graph of the trigonometric
functions is a tremendous help in understanding equations
and inequalities. A basic sketch can help reveal the number
of solutions in  and the quadrant of each solution.
For nonstandard angles, the value given by the inverse
function can then be used as a basis for stating the solu-
tion set for all real numbers. We’ll illustrate the process
using a few simple examples, then generalize our obser-
vations to solve more realistic applications. Consider the
function  a sine wave with amplitude 2, 
and a vertical translation of  To find intervals in 
where  we reason that f has a
maximum of  and a minimum of
since  With no phase
shift and a standard period of  we can easily draw a
quick sketch of f by vertically translating x-intercepts and
max/min points 1 unit up. After
drawing the line  (see
Figure 6.37), it appears there are
two intersections in the interval,
one in QI and one in QII. More
importantly, it is clear that
between these two
solutions. Substituting 2.5 for f(x)
in  we solve for sin x to obtain
which we use to state the solution in exact
form: for 
In approximate form the solution interval is
x  10.85, 2.292.
x  1sin
1
0.75,   sin
1
0.752.f 1x27 2.5
sin x  0.75,
f 1x2 2 sin x  1,
f 1x27 2.5
y  2.5
2,
1  sin x  1.1  2112 1,
3  2112 1
f 1x27 2.5,30, 22
 1.
f 1x2 2 sinx  1,
30, 22
If the function involves a horizontal shift, the graphi-
cal analysis will reveal which intervals should be chosen
to satisfy the given inequality.
Illustration 1
Given solve
for 
Solution
Plot the x-intercepts and maximum/minimum
values for a standard sine wave with amplitude 3, then shift 
these points  units to the right. Then shift each point one 
unit down and draw a sine wave
through the points (see Figure
6.38). This sketch along with the
graph of  is sufficient to
reveal that solutions to
occur in QI and QIII,
with solutions to 
outside this interval. Substituting
for  g(x) and isolat- 
ing the sine function we obtain 
then  after taking the inverse 
sine of both sides. This is the QI solution, with
being the solution in 
x  c  sin
1
a
1
15
bd
4
x  sin
1
a
1
15
b
4
sinax 
4
b
1
15
,
1.2
g1x21.2
g1x21.2
y 1.2
4
x  30, 22.g1x21.2
g1x2 3 sinax 
4
b 1,
Figure 6.37
Figure 6.38
y  2.5
f(x)
2
3
3
y
x
f(x)
2
4
3
y
x
 
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