MATH INVESTIGATIONS 4 Problem Set: 6 Fall’13
Teacher (circle): Condie Dover Krouse Pandya
Prince
NC
ID Number____________
Mods: __________
means you have to show all work done by hand, with no calculator use.
Find exact value for log(100!) – log(99!).
NC
1)
NC
2) Solve the following equation for x where 0  x  2 :
sin  x   3  cos  x    2
3) Given MAX with mA  36 , x  17, and a  13 , solve for the length of side m to the nearest
hundredth.
NC
4)
Solve each equation for n:
 2n 
 2n  1
a) 13  
  7

 n  2
 n3 
NC
5)
6)
NC
7)
 3n  1  3n  1
b) 4  


 n   n 1 
 a b 
If 
 rotates points in the plane though an angle  about the origin in the clockwise
b a 
direction, find tan  .
Use proof by Mathematical Induction to prove that for all n
,
n
k 
k 2
 
 2  
n(n  1)(n  1)
.
6
Find exact values for each of the following:
a)

3
 5 
cos  sin –1    tan –1   
5
 12  

PS 6.1
b)

1
 5 
sin  cos –1    sin –1    
3
 8 

Rev. F13
MATH INVESTIGATIONS 4 Problem Set: 6 Fall’13
Teacher (circle): Condie Dover Krouse Pandya
Prince
NC
ID Number____________
Mods: __________
8) Find the sum of each series, leaving complex answers in a + bi form. You may again assume
that for complex numbers a  bi, with a  bi  1 our formula for the sum of infinite geometric
series is valid.

i
a)   
n 1  2 
1 i 
b)  

n 1  2 
n

n
9)
In triangle ABC, AB = BC, AC = 1, and A  72 . Find the area and the other sides of this
triangle.
10)
In triangle ABC, the angle at C is 50 , a = 12, and c = 10. Solve the triangle.
NC
11) Erik wants to form 4-digit integers using only the digits 1, 2, 3, and 4. He is allowed to
repeat digits. What is the probability that he forms a 4-digit integer where at least one digit
repeats? Express your answer as a common fraction.
NC
12) Suppose that an isosceles triangle has two sides of length a and one side of length c.
a. Find the area of the triangle in terms of a and c. Simplify as much as you can.
b. A well-known formula for the area of a triangle is called Hero’s Formula. It is given by
s(s  a)(s  b)(s  c), where s 
abc
.
2
Use this formula to find the area of the isosceles triangle from part a.
Notes: 1. You may not use this formula in part a.
PS 6.2
Rev. F13
MATH INVESTIGATIONS 4 Problem Set: 6 Fall’13
Teacher (circle): Condie Dover Krouse Pandya
Prince
ID Number____________
Mods: __________
1)
2)
 100! 
log(100!)  log(99!)  log 

 99! 
 log 100 
2
sin  x   3  cos  x    2 
1
3
2
 sin  x  
 cos  x   

2
2
2

2

sin  x    

3
2

 5
 2k 
  4
x 
,
3  7
 2k 
 4
11
17 
x
or
12
12
3.
k

4a)
 2n 
 2n  1
13  
  7

 n  2
 n3 
13 
(2n)!
(2n  1)!
 7

(2n  n  2)!(n  2)!
(2n  n  2)!(n  3)!
13 7  (2n  1)


1
(n  3)
Using Law of Cosines:
132  m2  172  2  m  17cos(36 ) 
m2  27.506m  120  0 
m  5.44 or m  22.07
13n  39  14n  7 
n  32
(Note: Law of Sines should also lead to two
possible solutions)
PS 6.3
Rev. F13
MATH INVESTIGATIONS 4 Problem Set: 6 Fall’13
Teacher (circle): Condie Dover Krouse Pandya
Prince
4b)
5)
 3n  1  3n  1
4


 n   n 1 
Solution:
4
(3n  1)!
(3n  1)!


(3n  1  n)!n! (3n  1  n  1)!( n  1)!
4
(3n  1)!
(3n  1)!


(2n  1)!n! (2n)!( n  1)!
ID Number____________
Mods: __________
 a b   cos  sin  
b
b a     sin  cos   tan   a

 

8n(n  1)  (3n  1)(3n) 
n 2  5n  0 
n(n  5)  0  n  0 or n  5
3
 5
7a) Let   sin –1   and   tan –1   , so we
5
 12 
6) To be graded by teacher.
Proof(by induction):
Base Case (n=2):
have the following triangles:
2(2  1)(2  1)
 2
 1; RHS 
1

6
 2
LHS  
Inductive Step: Assume that for some m,
 k  m( m  1)( m  1)
. Then,
 
k 2  2 
6
m

m 1  k 
 2  3
  


k 2  2   2   2 




cos  sin

 m   m  1


2  2 
m( m  1)( m  1)  m  1


6
 2 

m( m  1)( m  1)

6
–1  5  
   tan     cos     
5
 12  
–1  3 
 cos    cos    sin    sin  
4 12

5 13

( m  1) m


3 5

5 13
33
65
2
 ( m 1) 1 
 m ( m  1) 
 
2
 6

( m  1)(( m  1)  1)(( m  1)  1)
6
So, by Mathematical Induction, the statement is true for all
natural numbers n > 1
PS 6.4
Rev. F13
MATH INVESTIGATIONS 4 Problem Set: 6 Fall’13
Teacher (circle): Condie Dover Krouse Pandya
Prince
1
 5
7b)Let   cos –1   and   sin –1    so we
3
 8
8a)
i
n
i
 
2

  
i
n 1  2 
1
2
i

2i
i (2  i )

(2  i )(2  i )
1  2i
1 2

or
 i
5
5 5
have the following triangles:

sin  cos


1  5  
   sin      sin     
3
 8 
–1  1 
 sin    cos    sin   cos   
2 2 39
5 1


    
3
8
8 3



ID Number____________
Mods: __________
2 78  5
24
Note:
8b)
i 1
 1
2 4
9)
1 i
n
1

i


2


 
1
i
n 1  2 
1
2
1 i

2  (1  i )
(1  i )(1  i )

(1  i )(1  i )
2i

= i
2

1
Then area  K  c 2 sin(36 )  .7694
2
1 i
1
1 1
     
1
2
2
2 2
2
Note:
c
1

 c  1.618
sin(72 ) sin(36 )
2
PS 6.5
Rev. F13
MATH INVESTIGATIONS 4 Problem Set: 6 Fall’13
Teacher (circle): Condie Dover Krouse Pandya
Prince
ID Number____________
Mods: __________
11) There are 4  4  4  4  256 possible numbers
10)
allowing for repeated digits.
There are 4  3  2 1  24 ways to arrange the digits
1,2,3,4 without repeating.
So probability of repeating at least one digit is:
12
sin( A)
10

 A  66.8 or 180  66.8  113.2
P
sin(50 )
256  24 232 116 58 29




256
256 128 64 32
 B  63.2 or B  16.8 .
Then Area 
1
 12  10 sin( B )  53.6 or 17.3
2
Triangle 1: A  66.8 , B  63.2, K  53.6
Triangle 2: A  113.2 , B  16.8, K  17.3
12a)
12b) Using the triangle in 12a) we see
s
a  b  c 2a  c

. Then,
2
2
K  s ( s  a )( s  b)( s  c )
 s ( s  a )( s  a )( s  c)
2
 2a  c  2a  c
 2a  c
 2a  c

 
 a 
 a 
 c

 2  2
 2
 2

c
c
a     h2  h  a 2 
4
2
2
2
2
1
1
c
1
or c 4a 2  c 2
Then Area  ch  c a 2 
2
2
4
4
PS 6.6
c  c  c 
c

  a     a  
2  2  2 
2

1 
c2 
1
 c  a 2   or c 4a 2  c 2
2 
4
4
Rev. F13
MATH INVESTIGATIONS 4 Problem Set: 6 Fall’13
Teacher (circle): Condie Dover Krouse Pandya
Prince
PS 6.7
ID Number____________
Mods: __________
Rev. F13