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MAT 182 – ALGEBRA & TRIGONOMETRY II
CHAPTER 7 SAMPLE TEST
1.
Convert to radian measure. Express your answer as a multiple of .
a.
2.
250°
3
5
b.
– 12

If s denotes the length of the arc of a circle of radius r subtended by the
central angle , find the missing quantity. Round your answer to three
decimal places.
a.
4.
b.
Convert to degree measure.
a.
3.
–135°
r = 3 ft, s = 8 ft,  = ?
b.
 = 120°, r = 2 m, s = ?
A denotes the area of the sector of a circle of radius r formed by the central
angle . Find the missing quantity. Round answers to three decimal places.
a.
 = 2 rad, r = 3 ft, A = ?
b.
A = 2 ft2, r = 5 ft,  = ?
5.
Convert 25.206° to an angle in D° M' S" form.
6.
A point on the terminal side of an angle, , is (2, –3). Find the exact value of
each of the six trigonometric functions.
7.
If sin  = – 3 and 90 <  < 270, find the exact value of each of the remaining
2
trigonometric functions.
8.
4
If cot  = 3 and cos  < 0, find the exact value of each of the remaining
trigonometric functions of .
9.
10.
Find the EXACT values of the following. Do not use a calculator.
csc (30)
b.
cos
c.
tan (21)
d.
csc 420°
e.
f.
sec 300
g.
3
sin 2
2
cot 3
h.
sin 4
i.
tan 225
j.
cos –
k.
cot (–45)
l.
tan
m.
sin (–3)
n.
csc
c.
e.
12.

5
6
3
2
9
2
Use your calculator to approximate the following to two decimal places.
a.
11.
5
6
a.
sin (47 52' 3")
4
csc – 5 
sec (–25.67)
8
7
g.
cot
i.
sin 3
b.
sec 2.8
d.
tan 7
f.
csc (–1.9)
h.
cos 303.52
j.
csc (–352°)
Find the exact values of the following expressions. Do not use a calculator.
a.
sin 45 cos 45 + tan 60 cot 30 + sec 60 csc 60
b.
1 + tan2 6 – csc2 4
c.
3 sin

2
3
– 4 cos

5
2
Graph each over at least one period. Make sure you state the period and
phase shift.
a.

y = tan 3 x
b.
1
y = 2 cos x + 1
c.
y = sin (2x – )
d.
y = csc 2x
e.
y = – sec 4x
f.
y = 2 cos (x + 6)
1

13.
Write the equation of a sine function that has amplitude = 2 and
period: 3.
14.
Find an equation of the graph below.
2
–2
2
4
CHAPTER 7 SAMPLE TEST ANSWERS
1.
3
4
a.
– 4
cos  = – 5
b.
25
18
tan  = 4
3
5
2.
a.
b.
108
–15
a.
8
 2.667
3
4
 4.189
3
csc  = – 3
5
sec  = –4
4
3.
b.
4.
5.
6.
a.
b.
cot  = 3
9.
9 sq ft
4
= 0.16
25
25 12' 22"
sin  = –
3 13
13
sec  =
cot  =
7.
13
3
13
2
2
–3
2
sin  = – 3
5
3
cos  = –
tan  =
csc  =
2 5
5
3
–2
sec  = –
cot  =
8.
3 5
5
5
2
3
sin  = – 5
2
b.
–
c.
0
d.
2 3
3
e.
f.
–1
2
g.
–
h.
2 13
cos  = 13
3
tan  = – 2
csc  = –
a.
10.
3
2
3
3
2
2
i.
1
j.
–
k.
l.
m.
n.
–1
undefined
0
1
a.
b.
c.
d.
e.
f.
g.
h.
i.
j.
0.74
–1.06
–1.70
0
1.11
–1.06
2.08
0.55
0.14
7.19
3
2
11.
12.
+4
3
7
2
b.
21 + 8 3
6
2
–3
c.
3 3
2
a.
y = tan 3x
3
y = sin (2x – )
p=
c.
a.
=

ps = 2
1

p=3
no ps
–
d.
–
3
2
3
2

2

2
–1
y = csc 2x
p=1
no ps
1
3
4
1
b.
y = 2 cos x + 1
p = 2
no ps
–1
1
4
1
2
3
2
1
e.
y = – sec 4x

1
2
p=2
no ps

2

3
2
2

2
1
–1

4

1
f.
y = 2 cos (x + 6)
y = 2 sin 3x
14.
y = 2 cos 2x
p = 2

ps = – 6
2
13.
1
other possible answers:

1
y = –2 sin (2x – 2)
5
6

–
6
–1
1

y = 2 sin (2x + 2)
11
6
etc.