Download Alpha Trigonometry Solutions 2013 FAMAT State

Alpha Trigonometry Solutions
2013 FAMAT State Convention

5
2
2

  

1. sin
. arcsin  
   since the range of the arcsine function is   ,  . A
4
2
4
 2 2
 2 
2. If  is an acute angle in a right triangle then the length of the adjacent side is 5 while the
length of the hypotenuse is 8. The length of the opposite side is therefore 39 . The cotangent
of  is the length of the adjacent side divided by the length of the opposite side, and hence
5
5 39
. B
cot 

39
39
3. The area of the triangle will be maximized when the side of length 2 is the smallest of the
three sides. This will occur when this side is the side opposite the angle of 30 . The length of
1
the other leg is therefore 2 3 and hence the area is A   2  2 3  2 3 . E
2
1

1


4. y  csc  2 x   
will have a vertical asymptote whenever

2
3


2sin  2 x  
3


k 


.
sin  2 x    0 . This will occur when 2 x   k , where k is an integer. Then x 
3

3

5
Of the choices, this will only occur when x 
. D
6
3
5. 1  sec 2   tan 2    sec   tan   sec   tan     sec   tan   and hence
2
2
sec   tan   . B
3
 


3


6. If 2 sin  2 x    3 , then sin  2 x   
. Either 2 x    2 k and hence
6 3
6
6 2



 2

x    k or 2 x  
 2 k and hence x    k . The solutions in 0, 2  are
12
6
3
4
  13 5   13 5 1  3  13  15  32 8
,
 




therefore x  , ,
.
. D
12 4 12 4 12 4 12
4
12
12
3
3


7. If the terminal side of 6 lies in the third quadrant, then   6 
and hence    .
2
6
4
1
2
 0.707 . The only choice in this interval is 0.59. C
Then  sin  
2
2
2013
2013
i
 i 


2013
2013

8.  cos  i sin    e 6   e 6  cos
 i sin

6
6
6
6

 
2013
2013
3
3
cos
 i sin
 cos
 i sin
 i . A
6
6
2
2
13
20
13
1
400 20
 2 


9. 213'20'' is equal to 2  
degrees. Converting this to
60 3600
60 180 180 9
20 


 . B
radians,
9 180 81
10. The graphs of y  cos x and y  x intersect at only one point. B
2013
Page 1 of 3
Alpha Trigonometry Solutions
11. The largest angle is the angle opposite the side of length
 
 
2013 FAMAT State Convention
5 . Using the Law of Cosines,
2
and hence C  135 . C
2
12. The point P  x, y  on the unit circle is represented in terms of  as  cos , sin   , so
5
2
 12 
2
2
 2 1 2 cos C . cos C  
y  sin  . A
13. is perpendicular to the radius of the circle, which has length 1. The radius and the line
segment PQ therefore form the legs of a right triangle. tan  
PQ
1
and hence the length of
PQ is given by tan  . D
14. x  r cos and y  r sin  , where r 2  x 2  y 2 . We than have that r 2  4  12  16 and
y
hence r  4 . Notice that tan    3 . Noticing that 2, 2 3 is in the third quadrant, we
x
have   240 . The polar coordinates are therefore  4, 240 . E


15. The horizontal asymptotes of y  arctan x correspond to the vertical asymptotes of

  
y  tan x whose domain is restricted to   ,  . These vertical asymptotes are x   and
2
 2 2


x  . y  is therefore a horizontal asymptote. D
2
2
 
 


16. The range of sin  3 x   is [1,1] , so the largest 4sin  3 x   can be is 4. The largest
12 
12 


 

possible value of f  x   4sin  3 x    2 is therefore 2. C.
12 

sin    cos  
 sin 2   2sin  cos  cos2   sin 2   sin  2   cos2  . Since
2
2
5
sin  2   and sin 2   cos 2   1 ,  sin    cos     . Therefore
3
3
5
15
. However, since  is a first quadrant angle, sin and cos must be
sin  cos 

3
3
positive and hence we disregard the negative solution. B
2
18. If tan    , the angle  is in an acute triangle with an opposite side of length 2 and an
7
adjacent side of length 7, and hence a hypotenuse of length 53 . The secant of this angle would
17.
2
53
, however since  is a second quadrant angle, the secant of  will be negative. A
7
19. If two angles of the triangle 60 and 45 , then the third is 180  60  45  75 . The
longest side will be opposite the largest angle and the shortest side will be opposite the smallest
2
c

angle. By the Law of Sines
, and hence c  2 2 sin 75 . Notice that
sin 45 sin 75
2 6
.
75  30  45 , so sin 75  sin  30  45   sin 30 cos 45  sin 45 cos 30 
4
Therefore c  1  3 . C
be
Page 2 of 3
Alpha Trigonometry Solutions
20.
89
89
k 1
k 1
sin k 
2013 FAMAT State Convention
89
 tan k    cos k   
k 1
cos  90  k   cos89 cos88


cos k 
cos1 cos 2
21. sec2   3sec  2   sec  2sec 1  0 .
sec 1  0    0 .


cos 2 cos1

1. D
cos88 cos89
 sec  2   0   

3
and
B
1 1
 . B
3
3
2 2
1  cos 
 
 2sin 2   . D
23. versin  1  cos   2
2
2
22. versin
24.
 1  cos
 1
 csc2   tan 2   1
 csc2   sec2  
cot  csc  tan  sin   cos 

cos


cos





2
2
csc2 
csc

csc





cos  1  tan 2    cos  sec 2   sec  . D


 16 . As  varies the curve is therefore a circle of
2
2
radius 4 starting at  4, 0  moving counterclockwise to the point  4,0  . The curve is therefore
25. Notice that x 2  y 2  16 cos 2
 16 sin 2
half of a circle of radius 4 and hence the arc length is 4 . C
u v
26. Let  be the acute angle u and v. Then cos  
. Since u  v  2 , u  0 , and
u v

3
. u lies in the fourth quadrant.
v  0 we must have that cos  0 . As a result,   
2
2
Therefore v can lie in either the first, second, or third quadrants but not the fourth. D
2
2 3
 4x 

27. The period of 7sin   is
while the period of 2cos  9x  is
. The period of
4
9
2
 3 
3
the sum of these two functions will be the least common multiple of these individual periods.
 4x 
Since the denominators of the periods are relatively prime, the period of 7sin    2cos  9 x 
 3 
will occur at an integer multiple of  . This will occur at 6 . C
28. The range of the arcsine function is the domain of the (restricted) sine function. The
  
restricted sine function has domain   ,  , and hence the domain of f ( x)  arcsin x is also
 2 2
  
  2 , 2  . E
 3

1


cos x  sin x   2 3 cos   x  , the maximum
29. Notice that 3cos x  3 sin x  2 3 
2
6

 2

value of which is 2 3 . C
30. Since the cosecant has a period of 360 ,
csc  2130  csc  2130  6  360  csc  30  2 . A
Page 3 of 3