Exam 2, version A
MATH 1060
Solutions
1. (10 points)
(a) Convert 50◦ to radians.
(b) Convert −512◦ to radians.
(c) Convert
π
7
radians to degrees.
(d) Convert 4 radians to degrees.
Solution:
Use the relationship π radians = 180◦ (or 2π radians = 360◦ ).
(a) 50◦ = 50◦ ×
π
180◦
radians.
(b) −512◦ = −512◦ ×
(c)
π
7
=
π
7
×
(d) 4 = 4 ×
180◦
π
180◦
π
π
180◦
radians.
= ( 180
)◦
7
= ( 720
)◦
π
2. (10 points) Find the following values:
(a) sin(30◦ )
(b) cos π4
(c) tan(120◦ )
(d) sec 17π
6
Solution:
(a) sin(30◦ ) =
1
2
1
Fall 2008
(b) cos π4 =
√
2
2
(c) tan(120◦ ) The reference angle for 120◦ is 180◦ − 120◦ = 60◦ .
√
◦
sin(60 )
=
tan(60 ) =
cos(60◦ )
◦
3
2
1
2
=
√
3
Since 120◦ is in quadrant 2 the y-coordinate stays the same and the x-coordinate becomes
negative, so
√
tan(120◦ ) = − tan(60◦ ) = − 3
(d)
sec(
The reference angle for
√
− cos( π6 ) = − 23 . So
17π
1
1
)=
17π =
6
cos( 6 )
cos(2π +
5π
6
is
sec(
π
.
6
cos( π6 ) =
5π
)
6
√
3
,
2
so since
=
5π
6
1
cos( 5π
)
6
is in quadrant 2, cos( 5π
)=
6
17π
1
2
1
√ = −√
)=
5π =
6
cos( 6 )
3
− 23
3. (20 points) Consider the right triangle:
(a) Find side length c if a = 4 and b = 10.
(b) Find sin A if c = 5 and b = 3.
2
(c) Find the length of the hypotenuse if tan A = 4 and a = 1.
(d) Find b if cos B =
1
2
and c = 1.
Solution:
(a) Find side length c if a = 4 and b = 10.
The Pythagorean Theorem tells us a2 + b2 = c2 , so c =
√
a2 + b2 =
√
16 + 100 =
√
116.
(b) Find sin A if c = 5 and b = 3.
sin =
OPP
HYP
so
sin A =
a
c
We need to know side a.
√
√
√
a2 + b2 = c2 , so a = c2 − b2 = 25 − 9 = 16 = 4.
Thus sin A =
4
5
(c) Find the length of the hypotenuse if tan A = 4 and a = 1. The hypotnenuse is c, and
√
√
c = a2 + b2 = 1 + b2 , so we need to find b.
tan =
OPP
ADJ
so
a
1
=
b
b
tan A =
but tan A = 4, so
1
b
Now
(d) Find b if cos B =
= tan A = 4, so b =
√
1
2
1
4
r
r
r
1 2
1
17
c=
+ = 1+( ) = 1+
=
4
16
16
√
√
and c = 1. b = c2 − a2 = 1 − a2 , so we need to find a.
a2
b2
cos =
so
cos B =
but cos B = 21 , so a = 21 .
q
q
√
So b = 1 − a2 = 1 − ( 12 )2 = 1 −
1
4
=
3
ADJ
HYP
a
a
= =a
c
1
q
3
4
√
=
3
2
4. (10 points) Compute the following values:
(a) arcsin(sin(30◦ ))
(b) arccos(cos π)
(c) arctan(1)
(d) arcsin(sin θ) where
π
2
< θ < π (Your answer for this one will be an expression in θ.)
Solution:
(a)
1
π
arcsin(sin(30◦ )) = arcsin( ) =
2
6
(b)
arccos(cos π) = arccos(−1) = π
(c) arctan(1) = θ, where − π2 < θ <
√
cos π4 = sin π4 = 22 .
π
2
is the angle such that tan θ = 1. This angle is π4 , since
So arctan(1) = π4 .
(d) arcsin(sin θ) where
π
2
< θ < π (Your answer for this one will be an expression in θ.)
arcsin takes a height as input and outputs an angle between − π2 and π2 whose sin is
that height. If one were to draw a horizontal line at the input height, the output of
arcsin for that height would be the angle where that line crosses the right semi-circle. In
this problem, since θ is in the second quadrant, the angle where the line at height sin θ
crosses the right semi-circle is at π − θ.
arcsin(sin θ) = π − θ
5. (10 points) Express the following in terms of sin θ and cos θ. Assume θ is such that
the indicated quantity is defined.
(a) sin(2π + θ)
(b) sin(π + θ)
4
(c) cos(π − θ)
(d) tan(−θ)
Solution:
(a) sin(2π + θ) = sin θ since sine is 2π-periodic.
(b) sin(π + θ)
The point corresponding to angle π + θ on the unit circle is antipodal (directly opposite)
the point corresponding to angle θ. The coordinates of one point are the negatives of
the coordinates of its antipodal point, so
sin(θ + π) = − sin θ
(c) cos(π − θ)
cos(π − θ) = cos(−θ + π). Just like in the previous part, −θ + π and −θ are antipodal,
so
cos(−θ + π) = − cos(−θ)
Cosine is even, so cos(−θ) = cos θ, so
cos(π − θ) = − cos(−θ) = − cos θ
(d)
tan(−θ) =
sin(−θ)
− sin θ
=
cos(−θ)
cos θ
6. (10 points) Are the following functions even, odd or neither?
(a) sin θ
(b) cos θ
(c) tan θ
(d) (sin θ) · (cos θ) − cos θ
Solution:
5
(a) Sine is odd. Sine tells you height, and if you go to angle −θ you will the negative of the
height at angle θ.
(b) Cosine is even. Cosine tells you width, and width doesn’t change if you go to anlge θ or
−θ.
(c) Tangent is odd.
tan(−θ) =
sin(−θ)
− sin θ
=
= − tan θ
cos(−θ)
cos θ
(d) (sin θ)·(cos θ)−cos θ is neither. Lets name this expession f (θ), so f (θ) = (sin θ)·(cos θ)−
cos θ
f (−θ) = (sin(−θ)) · (cos(−θ)) − cos(−θ) = (− sin θ) · (cos θ) − cos θ
It does not appear the the right hand side is equal to either −f (θ) or f (θ), so we suspect
f (θ) is neither even nor odd. To know for sure we check some points.
f (π) = (sin(π)) · (cos(π)) − cos(π) = 0 · (−1) − (−1) = 1
f (−π) = (sin(−π)) · (cos(−π)) − cos(−π) = 0 · (−1) − (−1) = 1
This tells us that f (θ) is not odd, because f (−π) 6= −f (π).
π
π
π
π
f ( ) = (sin( )) · (cos( )) − cos( ) = 1 · (0) − (0) = 1
2
2
2
2
−π
−π
−π
−π
) = (sin(
)) · (cos(
)) − cos(
) = −1 · (0) − (0) = −1
f(
2
2
2
2
This tells us that f (θ) is not even, because f ( π2 ) 6= f ( −π
).
2
7. (5 points) Here is a graph of cosine. Sketch a graph of secant on the same axes.
6
1.0
Kp
3
Kp
2
Kp K
0.5
1.0
Solution:
sec =
Important features of the graph of secant are:
7
1
cos
p p
1.5
x
p
2
p
2.5
p
3
p
• Secant is positive when cosine is positive, is equal to 1 exactly when cosine is equal to
1, and is greater than 1 if cosine is positive and less than 1.
• Secant is negative when cosine is negative, is equal to -1 exactly when cosine is equal
to -1, and is less than -1 if cosine is negative and less than -1.
• When cosine is zero secant is undefined. In particular, if cosine is very close to zero
then secant has very large magnitude, so secant has vertical asymptotes at the zeros
of cosine.
4
y
Kp
3
Kp
2
Kp K
2
0.5
p0
0.5
K
K
2
4
8. (10 points)
8
p p
1.5
x
p
2
p
2.5
p
3
p
(a) What is the area of the shaded region?
(b) What is the length of the shaded arc?
(c) What is the area of the shaded region?
(d) What is the length of the shaded arc?
9
Solution:
(a) What is the area of the shaded region?
The area of the sector is
is πR2 = π.
1
2π
The sector has angle 1. The whole circle has angle 2π.
fraction of the area of the entire circle. The area of the circle
Therefore, the area of the sector is
1
2π
× π = 12 .
(b) What is the length of the shaded arc?
10
The length of the arc is
1
2π
The arc has angle 1. The whole circle has angle 2π.
the length of the entire circle, which is 2πR = 2π.
Therefore, the length of the arc is
1
2π
× 2π = 1.
(c) What is the area of the shaded region?
The area of the sector is
t
2π
The sector has angle t. The whole circle has angle 2π.
the area of the entire circle, which is πR2 .
Therefore, the area of the sector is
t
2π
× πR2 =
(d) What is the length of the shaded arc?
11
tR2
.
2
The length of the arc is
t
2π
The arc has angle t. The whole circle has angle 2π.
the length of the entire circle, which is 2πR.
Therefore, the length of the arc is
t
2π
× 2πR = tR.
9. (5 points) A wheel of radius 4 inches rolls along the floor at an angular speed of
1000 rpm (revolutions per minute). How fast (linear speed) is the wheel rolling in miles per
hour?
Solution:
The circumference of the wheel is 2π · 4 inches. Therefore, 1 revolution of the wheel in
angular movement is equal to 8π inches of linear movement.
1000 rpm =
1000 rev
1000 rev 8π in 60 min 1 ft 1 mile
480000π
=
×
×
×
×
=
mph ≈ 23.8 mph
min
min
1 rev
1 hr
12 in 5280 ft
63360
10. (10 points) You are erecting a broadcast antenna on level ground. You need to secure
the antenna with a wire. The wire should attach to the antenna 100 feet above the ground.
At the point where the wire attaches to the ground it should make an angle of 35◦ with the
ground. How long should the wire be?
Solution:
The wire forms the hypotenuse of a right triangle. The opposite side of the triangle is the
100 feet of the antenna.
OPP
sin =
HYP
so
100 ft
sin(35◦ ) =
HYP
12
HYP =
100 ft
≈ 174 ft
sin(35◦ )
No new questions beyond this point.
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