Mathematical Investigations: A Collaborative Approach to Understanding Precalculus
Name
Key
Trigonometry: Modeling the Seas
ANOTHER LOOK BACK
Let’s look again at the beginnings of trigonometry. Complete these problems and check your
work as necessary.
1.
Give the exact value of each of the following. Do these without using your calculator.
a.
3
 35 

cos 

 6 
2
b.
 19 
1
cot  
 4 
c.
3
 7 

sin 

 3 
2
d.

sec  
 2
e.
 44 
tan  
  3

3 
f.
 3 
csc    2
 4 
undefined
m0
2.
Given that cos( )  m,
1
1  m2

2
    , find each of the following in terms of m.
m
sin( ) 
1  m2
f.
cos(   ) 
m
b.
tan( ) 
1  m2
m
g.
cos(   ) 
m
c.
sec( ) 
1
m
h.
cos   
m
d.
csc( ) 
1
i.
 3

cos 
  
 2

1  m2
e.


sin     
2

j.
 3

sin 
  
 2

m
a.
1 m
2
m
© 2005 Illinois Mathematics and Science Academy®
Trig. 19.1
Re-Review
Rev. S05
Mathematical Investigations: A Collaborative Approach to Understanding Precalculus
Name
3.
Key
Solve for x:
a.
4 sin 3 (x)  2sin2 (x)  2sin(x)  1  0
 2sin x  1  2sin 2 x  1  0
sin x  
sin x   12
x
b.

7
 2k
6
 2k or x 
6
or
3sin(2x   )  1
x

4
c.
sin  2 x     13
2 x    sin 1  13 
2 1  tan 2 x   tan 2 x  3
x
csc(x)  sin(x)  2
1
sin x  sin x  2
e.

6
1
3
 k , k 
2sin2 (x)  1, if    x  3
sin 2 x 
1
2
1  sin 2 x  2sin x
sin x  
sin x  2sin x  1  0
x   4  k , k 
2
 sin x  1
2
0
sin x  1

 2k , k 
2


cos  2x    0.5, if 3  x  4

3
2 x  3   3  2k
x
f.
2sec2 (x)  tan 2 (x)  3
tan x  
x  2.972  k
1.741  k  x  2.972  k , k 
d.
k
, k
2
3 tan 2 x  1
2 x      sin 1  13 
and
x  1.741  k

1
2
2x 
2
3
 2 k
x  3  k

x
or
1
2

2
2
 3   3 
 4 ,  4 , 4 , 4 , 
x

 5 , 7 , 9 , 11 
 4 4 4 4 
2 x  2 k
x  k
10
3
© 2005 Illinois Mathematics and Science Academy®
Trig. 19.2
Re-Review
Rev. S05
Mathematical Investigations: A Collaborative Approach to Understanding Precalculus
Name
4.
Key
Find the equation of a function satisfying the following conditions:
 
 5

, 4 
a.
maximum at  , 2  , next minimum at 
3 
 6

2
5
2

1
period




period



B

2
2
6
6
2
vertical shift  12  2  4  1
amplitude 
1
2

 2  4  3
 
 
y  3cos  2  x     1
3 
 
4
range of 5, 5 , period 3, contains (, 0).
B  23  23
b.
answers
will
vary
5
6
3
5
2
 
y  5sin   x   
2 
3
 2
5
2

5
maximum at (0, 15), next maximum at (5, 15), range of 10,15
c.
vertical shift  12 15 10  52
amplitude 
5.

1
2
15  10 
B
2
period
y
25
 2
cos 
2
 5
15
25
2
2
5
 5
x 
 2
5
10
Give the amplitude, period, phase shift, maxima, minima, and zeros.


f (x)  2 sin  5x    1
g(x)  3cos 2 x  4 
a.
b.

3
 2sin  5  x  15    1
amplitude  2; period 
maxima:
5x  3 

x
3
2
 3cos  2  x  2  
2

; phase shift 
left
5
15
minima:
 2 k 
 5 x  3  2  2k 

7 2k

, y3
30
5
x

30


2 k
, y  1
5
k
zeros: 0  2sin  5 x 

3
  1  12  sin  5 x  3 

5 x  3  6  2k or 5 x  3 

x


30


5
6
 2k
2 k
 2 k
or x  
,k
5
10
5
© 2005 Illinois Mathematics and Science Academy®

amplitude  3; period  1; phase shift 
2

right
maxima:
minima:
 2 x  4  2k 
 2 x  4    2k 
4  2 k 2
 k
2

y  3 k  
x
x
  4  2 k 1 2
  k
2
2 
y  3
k  
zeros: 2 x  4  2  k

x
Trig. 19.3
Re-Review

2
 4  k 1 2 k
   ,k
2
4  2
Rev. S05
Mathematical Investigations: A Collaborative Approach to Understanding Precalculus
Name
6.
Key
The hum you hear on a radio when it is not tuned to a station is a sound wave of 60 cycles
per second. This is called the frequency.
a.
What is the period of the wave? (What would make sense here?)
1
period  second
60
The wavelength of a sound wave is defined to be the distance the wave travels in
one period. If sound travels at 1100 ft/sec, find the wavelength of a 60 cycle-persecond sound wave.
1100 feet 1
55
 second  feet  18.3 feet
second 60
3
b.
A building casts a shadow 200 feet long when the angle of elevation of the sun is 33o.
How tall is the building?
7.
h
tan 33  200
h  200tan33  129.88 feet
h
33
200 feet
8.
Two boats leave at the same time, on a rather dark night, from ports that are directly
opposite (east to west) of each other on a lake. The first boat travels at 28 mph on a
course of 110o [course is measure clockwise from true north]. The second boat travels on
a course of 212o. If the boats collide, what was the speed of the second boat?
212
110
20
58
h
h
28
h  28sin 20
sin 58  hx 
x
x
28 mph
9.
sin 20 
28sin 20
x
28sin 20
 11.29 mph
sin 58
Sketch. Mark the scale clearly.
a.
 x
b. g(x)  4 sec  
 2
f (x)  3sin1 (2x)
3
2
4
 12
2
1
2
4
2
 32
© 2005 Illinois Mathematics and Science Academy®
Trig. 19.4
Re-Review
Rev. S05