Download MTH 120 — Fall — 2007 Essex County College — Division of

MTH 120 — Fall — 2007
Essex County College — Division of Mathematics
Test # 31 — Created December 17, 2007
Name:
Signature:
Show all work clearly and in order, and box your final answers. Justify your answers algebraically whenever possible. You have at most 80 minutes to take this 100 point exam. No
cellular phones allowed. Don’t forget to sign your exam!
1. If the sec θ = 7 and 270◦ < θ < 360◦ , find the exact values of the following functions of θ.
You must clearly indicate the x, y and r.
Work: From the information given we can conclude that x = 1, r = 7, and that y < 0.
To find the value of y, solve
√
√
72 = 12 + y 2 ⇒ y = ± 48 = ±4 3.
√
So, y = −4 3
√
−4 3
(a) sin θ =
7
(b) cos θ =
2 points
1
7
2 points
√
(c) tan θ = −4 3
1
2 points
(d) cot θ =
−1
√
4 3
2 points
(e) csc θ =
−7
√
4 3
2 points
This document was prepared by Ron Bannon using LATEX 2ε .
1
2. Find and simplify completely
10 points
8
k
÷
9
k−1
.
Work:
8
k
÷
9
k−1
8!
9!
÷
k! (8 − k)! (k − 1)! (10 − k)!
(k − 1)! (10 − k)!
8!
·
k! (8 − k)!
9!
=
=
(10 − k) (9 − k)
9k
=
3. Determine the values of each of the six trigonometric functions for θ = 330◦ . You must
clearly indicate the x, y and r.
Work: x =
√
3, y = −1 and r = 2
√
1
sin 330 = −
2
cos 330 =
3
2
1
tan 330◦ = − √
3
csc 330◦ = −2
2
sec 330◦ = √
3
√
cot 330◦ = − 3
◦
◦
4. Write an expression for the nth term.
10 points
8
32
2, 1, , 1,
,...
9
25
Work: By inspection.
an =
2n
n2
5. Use your calculator to evaluate the trigonometric function. Round your answers to three
decimal places.
7π
(a) tan −
8
7π
Work: tan −
≈ 0.414
8
(b) cot 3.529
Work: cot 3.529 =
2 points
2 points
1
≈ 2.451
tan 3.529
2
(c) csc (−6.895)
2 points
Work: csc (−6.895) =
1
≈ −1.741
sin (−6.895)
(d) sec 60.001
2 points
Work: sec 60.001 =
1
≈ −1.050
cos 60.001
(e) arcsin (−0.645)
2 points
Work: arcsin (−0.645) ≈ −0.701 or −40.166◦
6. Find the term containing x3 in the binomial expansion of (3x − 2)10 .
Work:
10
3
10 points
(3x)3 (−2)7 = −414, 720x3
7. Solve for x. Restrict your solutions to [0, 2π).
10 points
2 cos2 x = 1 − cos x
Work:
2 cos2 x = 1 − cos x
2 cos2 x + cos x − 1 = 0
(2 cos x − 1) (cos x + 1) = 0
This results in two equations to solve.
cos x =
1
2
and
cos x = −1
As discussed in class, it is best to graph one cycle of the cosine function to visualize where
the solutions are. So here’s the graph of the cosine curve on the interval [0, 2π), with the
line segments, y = 1/2 and y = −1.
1
0
-1
Figure 1: One cycle of the cosine function, with line segments y = 1/2 and y = −1.
3
Clearly, we have
x=
π
5π
, π,
.
3
3
If you prefer degree measure, the answer is
x = 60◦ , 180◦ , 300◦ .
10 points
8. Find the first three terms in the expansion of
22
b−2/3 + b1/3
Work: Here’s what you need to do:
22 −2/3 22 1/3 0
22 −2/3 21 1/3 1
22 −2/3 20 1/3 2
b
b
+
b
b
+
b
b
.
21
20
22
Here’s the simplified form:
b−44/3 + 22b−41/3 + 231b−38/3 .
I will also accept the last three terms as an answer:
b22/3 + 22b19/3 + 231b16/3 .
5π
9. Find the exact value of arccos sin
.
3
Work:
√ !
5π
3
5π
arccos sin
= arccos −
=
3
2
6
10 points
or
150◦
10 points
10. A sequence is defined recursively by
10 points
an+2 = a2n − an+1 .
a1 = 1,
a2 = 2
find a5 .
Work:
a1 = 1
a2 = 2
a3 = a21 − a2 = 12 − 2 = −1
a4 = a22 − a3 = 22 − (−1) = 5
a5 = a23 − a4 = (−1)2 − 5 = −4
4
11. Graph at least one period (cycle) of
y = 2 sin
π
4
−
π x +3
4
6
5
4
3
2
1
-7
-6
-5
-4
-3
-2
-1
0
1
2
3
4
5
6
7
8
9
10
11
12
-1
-2
-3
Figure 2: Partial graph of y = 2 sin
π
4
−
π x + 3. I’m only looking for one cycle [black].
4
5