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Honors Math Analysis
Fall Mid-Term Exam
All numerical answers expressed to three places after the decimal (y.xxx).
1. The point (6, –12) is on the terminal side of an
angle in standard position. Give the smallest positive
angle in both degrees and radians.
12
1 12
tan  
   tan
 63.435
6
6
63.435  360  296.565  296.565

180
 5.176
2. Give the exact values of the six trigonometric
 19 
functions of the angle  
.
 6 
19
12 7
7
7
5



 2 

 2 
6
6
6
6
6
6
5 1
 19 
 19 
sin  
 ;csc  
  sin
2
6 2
 6 
 6 
 19 
cos  

 6 
 19 
tan  

 6 
3
 19
;sec  
2
 6
2


3

1
 19 
;cot  
 3
3
 6 
Honors Math Analysis
Fall Mid-Term Exam
3. State the (a) amplitude, (b) period, (c) phase shift,
(d) domain, and (e) range for the sinusoid


f  x   1.5sin  2 x   .
4

 

 

f  x   1.5sin  2 x    1.5sin  2  x   
4
8 

 
2

  ; phase shift =
2
8
Domain : x : x ; Range : y : 1.5  y  1.5
amplitude = 1.5, period =
4. Write g  x   2 cos 2 x  3sin 2 x as a sinusoid in
the form y  a sin  bx  c  .
y  a sin  bx  c   a sin bx cos c  a cos bx sin c
a cos x  2, a sin x  3, b  2
a 2  cos 2 x  sin 2 x   4  9  13  a  13
c  cos
1
2
 56.310
13
g  x   2 cos 2 x  3sin 2 x 
13 sin  2 x  56.310 
= 13 sin  2 x  0.93
Honors Math Analysis
Fall Mid-Term Exam
5. Find the length of the arc intercepted by a central
5
angle of
rad in a circle of radius 4.
4
5
s  r  4
 5
4
6. Dr. Thom Lawson standing on flat ground 62 ft
from the base of a Douglas fir measures the angle of
elevation to the top of the tree as 72º24′. What is the
height of the tree?
24
h

7224  72   72.4  tan 72.4 
60
62
h  62 tan 72.4  195.449 ft
7. What would be the angle of elevation of the top
of a 150 ft building as measured from a point 78 ft
from its base?
150
1 150
tan  
   tan
 62.526
78
78
8. If sin   0.843 and cos   0.537 , find sin 2θ.
sin 2  2sin  cos   2 0.843 0.537  0.905
Honors Math Analysis
Fall Mid-Term Exam
2
9. Prove : sin 3  3cos  sin   sin 3 
3cos 2  sin   sin 3   sin 3  sin  2   
 sin 2 cos   cos 2 sin 
 2sin  cos  cos   cos 2   sin 2   sin 
 2sin  cos 2   cos 2  sin   sin 3 
 3cos 2  sin   sin 3 
10. Use your calculator to conjecture whether
sec x  sin x tan x  cos x is likely to be an identity.
Confirm your conjecture algebraically.
sec x  sin x tan x  cos x is likely to be an identity.
1
sin x
cos x  sec x  sin x tan x 
 sin x
cos x
cos x
1  sin 2 x cos 2 x


 cos x
cos x
cos x
Honors Math Analysis
Fall Mid-Term Exam
11. Find and state all the solutions to
cos3 x  2sin x  0.7  0 in the interval [0, 2π)
graphically.
cos3 x  2sin x  0.7  0 at x  0.137,3.789
12. Two markers A and B on the same side of a
canyon are 80 ft apart, as shown in the figure. A
hiker is located on the opposite rim at point C. A
surveyor determines that BAC  70 and
ABC  65.
(a) What is the distance between the hiker and point
A?
C  180   70  65  45 
b
c

sin B sin C
c sin B 80sin 65
b

 102.537 ft
sin C
sin 45
Honors Math Analysis
Fall Mid-Term Exam
(b) What is the distance between the two canyon
rims? Assume they are parallel.
d
sin 70 
102.537
d  102.537 sin 70  96.353 ft
13. To determine the distance between two points A
and B on opposite sides of a lake, a surveyor chooses
a point C that is 900 ft from A and 225 ft from B, as
shown in the figure. If the measure of the angle at C
is 70º, find the distance from A to B.
c 2  a 2  b 2  2ab cos C
c
 225   900   2  225 900  cos 70
2
 849.769 ft
2
Honors Math Analysis
Fall Mid-Term Exam
14. Find the area of the triangle with sides 8, 9, 10 in
length.
8  9  10
s
 13.5
2
A  13.5 13.5  8 13.5  9 13.5  10 
 13.5  5.5  4.5  3.5   34.197
Honors Math Analysis
Fall Mid-Term Exam
15. Extra Credit Building inspector Julie Wong
checks a building in the shape of a regular octagon 20
ft on a side. She checks that the contractor has
located the corners of the foundation correctly by
measuring several diagonals. Calculate what the
lengths of HB, HD, and HC should be.
360
180  45
 45 
 67.5
8
2
HR
20
20sin 67.5

 HR 
 26.131
sin 67.5 sin 45
sin 45
HB 
 26.131   26.131  36.955 ft
2
2
26.131sin135
 48.284 ft
sin 22.5
HD  2 HR  52.263 ft
HC 