Download Mu Alpha Theta National Convention: Hawaii, 2005 Solutions

Mu Alpha Theta National Convention: Hawaii, 2005
Solutions – Trigonometry Topic Test – Mu Division
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1. Since  is an irrational number there are an infinite number of petals, by
definition.
sin  3 cos2  3
3 2 1 2
3
3 1 3 3
3  3
 (1) 




2.


3
2
3 2
6
6
3 3 1
 tan7 6 cot 7 4
3. The function in the neighborhood of x   2 is the line y=0 with the exception of
the point x   2, where y becomes 1. This is a removable discontinuity since

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 x with the nearly equal piecewise function y  sin x x   2
replacing y  sin
x  2
 0

allows y to be continuous at x   2.
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4. f (x)  24  cos(2 x) , f ( 2)  24  cos( )  24
5. 
Using the law of cosines at time t  2 , we get
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c 2  (4  2)2  (3 2)2  2  (4  2)  (3 2)  cos(60 )  64  36  48  52 , so the
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distance between the two points at t  2 is c  52 , since we are looking for the
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distance they are moving
 apart or dc dt , we shall differentiate the law of cosines,
with the only constant being  . This gives us an equation of
 db
dc
da
db 
da 
2c
 2a  2b  2cos
60 
a  b , plugging in for the various given


 dt
dt
dt
dt
dt 

and derived values 
gives us:
dc
2 2 13
 2(2  4)(4)  2(2  3)(3)  (2  4)(3)  (2  3)(4)
dt
dc
4 13
 52
dt
dc
 13
dt
sin( x)
1
du
dx , setting cos(x)=u becomes;  
6. 
, which is; ln u = ln =
cos( x)
u
u
ln sec( x)
 
 
7.
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4,  3,  6  1,
x  r cos sin   4  1 2  1 2
1
y  r sin  sin   4   3 2 1 2  3
r,,    x, y,z

z  r cos   4 
 3 2 2 3
3,2 3


lim 2tanh( 3x)  cosh( x)  sinh( x) =
x 


2   lim tanh( 3x) lim cosh( x)  sinh( x) =
x  
 x  
x
2  1  lim e  = 2
x 
9. f (x)  sin x sec x tan x  cos x csc x cot x  tan 2 x  cot 2 x 
sec 2 x 1 csc 2 x 1 sec 2 x  csc 2 x
8.
Mu Alpha Theta National Convention: Hawaii, 2005
Solutions – Trigonometry Topic Test – Mu Division
 f (x) dx   sec
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2
x  csc 2 xdx  tan x  cot x 
sin x cos x sin 2 x  cos2 x
1



 2csc(2x)
1
cos x sin x
sin x cos x
sin( 2x)
2

1
d
d 
10.
arcsin x  arccos x  arctan x    arctan x  2
 x 1
dx
dx 2

11. At x=0, we can see that dy/dx=1. This eliminates all answers except B and E,
dy
 cos x
noticing that the x interval is  2 , the rest of the values match up to
dx
  ln 2 ln 2
e i  ei
e
e
1 22 5


12. Since cos 
, we have cos(i  ln 2) 
4
2
2
2

1 2


13. r  sin  is a circle of diameter 1, centered at (.5, 0). The area is then    
2 
4
4
4
2
2
2
2
14.  cos x  sin xdx  

 (cos x  sin x)(cos x  sin x) dx
1
 cos(2x) dx  2 sin( 2x)  C


sin x  x 
cos x 1
sin x  1 sin x  1
15. lim 
 lim
 lim
 lim

 x 0 
 3x 2 
 x 0 
 6x 
 6 x 0 
 x 
 6
x 3 
x 0 
2
2
6
16. sin 6 (x)
  3sin (x)cos (x)  cos (x) 
sin 6 (x)  3sin 2 (x)cos2 (x)  sin 2 x  cos2 x cos6 (x) 
 
sin 6 (x)  3sin 4 (x)cos2 (x)  3sin 2 (x)cos4 (x)  cos6 (x)
3
sin 2 x  cos2 x   13  1
cos  3  h   cos 3
cosh   3  cos 3


17. lim 
 lim 


h
h



h0
h0 
cos 3  h  cos  3
. This equation to the left is now in the standard
lim 
h

h0
form for the limit form of the derivative. With f (x)  cos x at x    3.
3
Then f  3  sin  3  sin  3 
2
18. Define f (x)  tan x , then
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
Derivative f
@x=0
0
 Zero
tan x

2
First
1
sec x
 Second
2
0
2sec x tan x
2
2
4
Third
4 sec x tan x  2sec x 2

The 
third derivate term of the series will be divided by 3!, so the answer is 1/3.
19. The
parametric equations define an ellipse, the question is asking for the
perimeter
of the ellipse, it is well known that the perimeter of an ellipse cannot be

evaluated exactly. The integral if set up, is an elliptic integral which cannot be
solved precisely.
Mu Alpha Theta National Convention: Hawaii, 2005
Solutions – Trigonometry Topic Test – Mu Division
20. We know that cos(3 )  4 cos3   3cos , by double angle and addition of angles
formulas. Then plug in   20 , we get cos(60 )  4 cos3 (20 )  3cos(20 ) ,
simplifying and moving all terms to one side we get
0  8cos3 (20 )  6cos(20 ) 1, rewriting to get a polynomial equation in terms of

x and y, yieldsy  8x 3  6x 1, where x  cos(20 ) is a solution of the equation.
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21. For f (x)  sin x , f (14 ) (x)  f (144 3) (x)  f (x)  sin x
22. Set e y  (tan x)tan x , then
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ln(tan x)
y  tan x  ln(tan x) 


cot x
sec 2 x 




ln(tan x)
sin 2 x 
 tan x 

lim 

lim

lim



2
2 x tan x 
x0  cot x  x0 csc x x0
cos


y
lim [tan x]  0 , that is lim y  0  lim e  e0  1
x0
x0
x0
arctan x
dx
23. 
, and the integral
dx , set u  arctan x , then du 
1 x 2
1 x 2
1
1
becomes  u du  u2 , which when put back in terms of x is, arctan 2 (x)
2
2

x

11
sin(
33)

sin(11)
sin(
3x)

sin(
x)
24.
rewritten with
, gives us
, expanding the

sin( 3x) to be in terms of sin( x) , we get
3sin( x)  4sin 3 (x) sin( x)  3sin 2 (x)  4sin 4 (x)  3sin 2 x  4sin 2 (x)(1 cos2 x) 


2
2
3sin 2 (x)  4 sin 2 (x) 4 sin 2 (x)cos2 (x)
 sin (x)  sin (2x) , which when
plugging x  11back into the equation gets the answer of sin 2 (22)  sin 2 (11) .
25. For ease of explanation BC  a , mBAC  A and similarly for the other sides
and angles. We have A  B  C  180 , so C  45 , now using the law of sines we
4 2 2

4 6
sin( A) sin( C)
sin( 60 ) sin( 45 ) 
have




c



3
a
c
4
c
32

3
5
7
x
x
x

 is the Maclaurin expansion for sin( x)
26. Yes. x  
3! 5! 7!
2

 No. 1 cos x only equals sin( x) when sin( x)  0
ix
ix
e
e

 sin( x)
No.

2i
1
1 

 No.
, the + C as well as csc(x) not being defined
 cot x csc x dx csc x  C


everywhere keep this from being possible.
Only the first one holds true. 1
27. h (x)  f (g(x)) g(x)  3tan 2 x sec 2 x

2
2
h( 3)  3tan 2 ( 3)sec 2 ( 3)  3 3  2  36
 



Mu Alpha Theta National Convention: Hawaii, 2005
Solutions – Trigonometry Topic Test – Mu Division
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28. The maximum of a function of the form f (x)  asin x  bcos x is
a  4 and b  3, the maximum of the function is 5.
1
29. y  sin 2 x  cos2 x  sin 2 (2x), now
4


2
 2
 2



1
1
1
cos(4x)
1
1
2

 4 sin (2x) dx  4  2 dx  8 x  4 sin( 4 x)  16
0
0
0
e x  ex
e x  ex
30. sinh( x) 
, cosh( x) 
, therefore:
2
2
cosh 2 x  sinh 2 x  cosh(2x)

d
sinh x  cosh x
dx

d
cosh
 x  sinh x
dx
2sinh x cosh x  sinh( 2x)
cosh 2 x  sinh 2 x  1
and only II and IV are true.
a2  b2 , with