Download Solutions - York University

Assignment 11
York University MATH1710
Exercise 1
For each of the following determine the quadrant that the angle lies in. Convert the angle to
radians if it is expressed as degrees and convert it to degrees if it is expressed as radians:
(a) 150
Since 90 < 150 < 180, the angle 150 falls into quadrant 2.
r
d

(r  radians, d  degrees)
2 360
2 d
r
360
2 (150) 300 5
r


360
360
6
150 
5
radians
6
(b) -33630’
We can add 360 to this angle, and get -33630’+360=2430’. Since 0 < 2430’ < 90, this
angle falls into quadrant 1.
r
r
2 d
360
2 (336
360
33630'  
(c)
30
)
60

673
360
673
radians
360

radians
7
 

< , the angle radians falls into quadrant 1.
7 2
7
Since 0 <
d
d

7
360r
2
360

7  360  25.7
2
14
radians  25.7
(d) 4.2 radians
If we add 3  2 to 4.2 , we get 1.8 . Since
4.2 radians falls into quadrant 4.
360r
d
2
360(4.2 )
d
 756
2
4.2 radians  756
3
 1.8  2 , the angle
2
(e) 780
We can subtract 2x360 from 780, and we have 60. Since 0 < 60 < 90, the angle 60
falls into quadrant 1.
2 d
360
2 (780) 13
r

radians
360
3
r
(f) 2 radians
Since   2 

2
, the angle 2 radians falls into quadrant 2.
360r
2
360(2) 360
d

 114.6
2

2 radians  114.6
d
Exercise 2
Recall that for a unit circle with center O if given two points P and Q then the arc length from P
to Q equals the measurement of the angle POQ in radians. Suppose now that there is a larger
circle with radius r and center O. Let Q’ be the intersection of the ray OQ with the larger circle
and let P’ be the intersection of the ray OP with the larger circle. Then we can say comparing arc
lengths to the total circumferences that the fractions one gets must be the same. That is:
arclength(PQ) arclength(P'Q')

2
2 r
This means that arclength(P'Q ')  r  arclength(PQ) . With this in mind, here is the problem.
Given that the earth is roughly a sphere and that the radius is roughly 6378 kilometers, find the
distance between Johannesburg, South Africa and Jerusalem, Israel, assuming that the two cities
are on the same longitude (Johannesburg due south of Jerusalem) and that their latitudes are
respectively 2611’S and 3147’N.
Q’
3147’ (Jerusalem)
Q
O
(0,0)
P
(1,0)
P’
-2611’ (Johannesburg)
We can calculate the arc length on the unit circle from -2611’ to 3147’ by determining
what proportion of the total circle is covered by this arc, in radians, and then multiplying
that by the radius. The size of this arc (from Jerusalem to Johannesburg) is 3147’ – (–
2611’), or 5758’. Converting this to radians:
r
r
2 d
360
(r  radians, d  degrees)
2 (57
58
)
60
360

2 (
3478
)
60
360

3478
radians
10800
Now, if we multiply this by the radius of the earth,
3478
6378  6453km
10800
Therefore, the distance from Jerusalem to Johannesburg is approximately 6453km.
Exercise 3
Evaluate, when possible and without the use of a calculator the six trigonometric functions for the
angular values
Quad sin
cos
tan
csc
sec
cot
Q3 sign=–
Q3 sign=–
(a)
3
5

5

5

5
5

sin
4
5
4
cos
 sin


 cos
4
1

2
tan
5
4

4

sin
5
4
cos
1
csc
4
5
4
2
1

2
1
1


2
(b)


3/4
2
Q3/4 sign=–
 
sin   
 2
  sin
 
cos   
 2

 cos
2
0
 1

2
 
tan   
 2
 
 2 

 
cos   
 2
sin  

(c)
2

3
2
Q2 sign=+
2
sin
3

sin
3
3

2
Q2 sign=–
2
cos
3

  cos
3
1

2
1
0
Undefined
2
tan
3

sin
2
3
cos
2
3
3
2

1

2
 3
4
1

5
sin
4
1

1

2

2
 
csc   
 2

1
 
sin   
 2

1
 1
1
csc
2
3
1

2
sin
3
1

3
2
2

3
sec
4
1

5
cos
4
1

1

2

1
 
cos   
 2
1
0
Undefined

sec
4
1

tan
5
4
1
 1
1
2
 
sec   
 2

cot
2
3
1

2
cos
3
1

1

2
 2
 
cot   
 2
 
cos   
 2

 
sin   
 2

0
0
1
cot
2
3
1

2
tan
3
1

3
Exercise 4
Without the use of a calculator evaluate
(a) sin7
sin7  sin(3  2   )  sin   0
(b) cos
9
4
cos
(c) sin

9


2
 cos  2    cos 
4
4
4
2

4
3
sin

4
2 
2

3
 sin  2 
 sin
 sin 

3
3
3
3
2

Figure 1: The sun, the moon, the earth
Exercise 5
When the moon is half full, the angle earth-moon-sun EMS is a right angle. At the same time
the angle sun-earth-moon SEM is 89.85. If the distance from the earth to the moon is
approximately 384,403 km, what is the approximate distance from the earth to the sun?
EMS = 90
SEM = 89.85
EM =384,403
ES = ?
adjacent
cos 
hypoteneuse
cos89.85 
0.002618 
EM
ES
384,403
ES
384403
ES 
0.00261799
ES  146831298
The distance from the earth to the sun, rounded to 6 significant digits, is approximately
146,831,000 km.
Exercise 6
Recall that multiplying a function definition by a positive constant causes the graph to be
vertically stretched if the constant is greater than 1 and to be vertically shrunk if the constant is
less than one but greater than zero. Also recall that to horizontally stretch one needs to multiply
the variable by a number between zero and one and to horizontally shrink one needs to multiply
the variable by a number greater than one. Suppose that f (x)  a sin bx and g(x)  a cos bx
where a and b are positive real numbers. The number a is called the amplitude and the number
2
is called the period. In the case that x represents time, the period represents the amount of
b
time for a complete cycle to pass. The frequency on the other hand is the reciprocal of the period
and represents the number of cycles in a unit of time - say, cycles per second. Any sound causes
vibrations of the air which like ripples in a pond form a sine curve. In the case of concert pitch A,
this is a sound that represents 440 cycles per second.
(a) Using a graphing utility, graph f (x)  a sin bx in the cases
i. a = 1 and b = 2 (blue)
ii. a = 1 and b = ½ (green)
iii. a = 2 and b = 2 (red)
y
2
1
-7
-6
-5
-4
-3
-2
-1
1
2
3
4
5
6
7
x
-1
-2
(b) Why is the number
2
called the period?
b
The period is the time between successive instances of an event - in this case, cycles of
the sine wave. The period is inversely proportional to b, in f (x)  a sin bx , because b is
a multipler of the variable x, which stretches or shrinks the wave horizontally. As b
increases, the period becomes smaller. Since one cycle is defined as 2 radians, 2 is
in the numerator, and b, which is inversely proportional to the period, is in the
denominator. Thus one cycle would be completed in
2
radians.
b
Exercise 7
The important points for a cosine or sine graph are those at which the function has value zero,
where it has a minimum value and where it has a maximum value. We know that the sine
function on the interval [0,2 ] has a minimum at x  0 , a max at x 
at x 
3
2

2
, a zero at x   , a min
and a zero at x  2 . The graph of the sine goes through a single basic cycle over the
interval [0,2 ] and this behavior repeats over all subsequent intervals of length 2 . The cosine
graph has similar but different behavior. On the interval [0,2 ] , it starts with a maximum value
at x  0 . If you are given for instance y  a cos(bx  c) or y  asin(bx  c) , then as before
the period is
2
, for b > 0, and the amplitude is |a|. The problem is to find where the basic cycle
b
starts and finishes. We know that the number c causes a horizontal translation. Here is what you
do. Look at the equations bx  c  0 and bx  c  2 and solve for x. The solutions give the
beginning and ending points of the basic cycle.
Here are the problems.


x  
For each of y  2sin  x  , and y  3cos    ,
3
2 2


Starting
point of the
basic cycle
bx  c  0
bx  c  0
x
x
Ending
point of the
basic cycle

0
(c)
x
3
x 1 0

 2
a2
2
sketch a
 4 7
, ,...
graph of two Zeros: ,
3
3
3
periods of
5 17 29
the function
Maximums:
,
,
,...
showing
6 6
6
zeros and
11 23 33
maximum
Minimums:
,
,
,...
6
6
6
and

2


bx  x  2
x
2
0
2 2
x    0
3
3
6  7
x
 
3 3
3
(b) Amplitude
Period

y  3cos   x   
3
(a)

y  2sin  x  
x  1
bx  c  2
x

 2
2 2
 x    4

x 1 4
x3
a3
2 2 2

4

1 
2
Zeros: 0,2,4,...
Maximums: 1,3,7,...
Minimums: 1,5,9,...
minimum
points
(d) verify (c)
with a graph
using a
graphing
utility.
y
2
y
3
2
1
1
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
x
-1
1
-1
-1
-2
-2
-3
2
3
4
5
6
7
x