Jim Lambers
Math 1B
Fall Quarter 2004-05
Pratie Quiz 5
No alulators of any kind are p ermitted for this quiz.
1. Compute sin
1
(
1
=
2).
Solution
Let
x
= sin
1
(
1
=
2). Then, by the denition of the inverse sine funtion, sin
x
=
1
=
2, where
=
2
x
=
2. Sine sin(
=
6) = 1
=
2, and sin(
x
) =
sin
x
, it follows that
sin(
=
6) =
1
=
2, and therefore sin
1
(
1
=
2) =
=
6.
2. Compute os
1
(os(
=
4)).
Solution
First, we ompute os (
=
4). We an use the fat that os(
=
4) =
p
2
=
2 and
the fat that os(
x
) = os
x
to onlude that os(
=
4) =
p
2
=
2 as well. Now, let
x
=
os
1
(
p
2
=
2). Then, by the denition of the inverse osine funtion, os
x
=
p
2
=
2, where
0
x
. The only angle in this interval that satises this equation is
=
4, so we onlude
that os
1
(os(
=
4)) =
=
4.
3. Given
y
= 1
3 os(4
x
2
), state the amplitude, perio d, and phase shift.
Solution
This funtion ts the form
y
=
k
+
A
os(
Bx
+
C
), where
k
= 1,
A
=
3,
B
= 4,
and
C
=
2
. It follows that the amplitude is
j
A
j
= 3, the p erio d is 2
=B
= 2
=
4 =
=
2,
and the phase shift is
C=B
=
(
2
)
=
4 =
=
2.
4. Sketh the graph of the urve
y
= 1
3 os(4
x
2
) from problem 3 from
x
on
the graph given below. Y
ou may use the seond sheet of this quiz for srath paper, but your
nal graph must be drawn b elow.
Solution
The graph is shown in the spae provided. To obtain this graph, the graph of os
x
is rst ontrated horizontally by a fator of
B
= 4, so that its period is 2
=B
=
=
2. Then,
the graph is shifted to the right by the phase shift of
C=B
=
=
2. Then, it is strethed
vertially by a fator equal to the amplitude of
j
A
j
= 3, and reeted aross the
x
-axis b eause
A
=
3
<
0. Finally, it is shifted up by one
k
= 1 unit.
In the graph shown, the bold portion of the urve orresponds to the graph of
y
= os
x
for 0
x
2
, after these transformations have been applied. One an pro due this graph
by applying these transformations to suh a portion of the graph, and then lling in the
remainder of the graph using the fat that osine is periodi.
1
