MSLC – Math 1149
Exam 2 Review
Disclaimer: This should NOT be used as your only guide for what to study.
1. Find the exact value of each of the following.

 99  
a. tan  sin 1  

 101  


 4  
b. cos 1  cos 

 3 

2. Given triangle ABC with the following properties:
b  14, c  17, B  44 and C is an obtuse angle
Find the measure of angle C.
Round your answer to 2 decimal places.
3. A team of surveyors have been hired to measure the distance across a canyon. Using a tree at
point T on the opposite site of the canyon as a reference point, they established points A, B, and
C and found the following distances:
AB = 12.25 ft
BC = 6.5 ft
AC = 15 ft
a. Find the measure of angle ABC.
Round your answer to 2 decimal places.
b. Find the distance TC across the
canyon to the nearest foot.
4. The predator population of a predator/prey model is modeled by the function:
P  t   1300cos3t  4000
where P is the predator population after t years.
a. What is the maximum predator population?
b. What is the minimum predator population?
c. Find the length of time between successive periods of maximum population.
Round your answer to 2 decimal places.
5. Given cot x  
a. sin 2x
x
b. sin
2
c. cos 2x
x
d. cos
2
2
and sin x  0 find:
3
6. Find the exact solution of:
11
a. sin
12
b. tan(165)
7
c. cos
8
7. Find all solutions of:
a. 2 cos 2 x  1  0 on the interval  0, 2 
b. cos x sin x  2cos x  0
x
c. 2sin  3  0 on the interval 0,8 
2
8. Establish the identity:
1  sin x
2
a.
  sec x  tan x 
1  sin x
sec  csc
b.
 sin   cos 
tan   cot 
cot x cot y  1
c. cot( x  y) 
cot x  cot y




d. sin   x   sin   x 
2

2

1  sin 2 x
1
e.
 1  sec x csc x
sin 2 x
2
ANSWERS
1. a. 
99
20
b.
2
3
2. C  122.49
3. a. ABC  101.84
4. a. 5300
b. 2700
5. a. 
12
13
6. a.
6 2
4
7. a.
b. TC  31 feet
b.
1
1

2
13
b.
c. 2.09 years
c. 
1 3
1 3
3 5 11 13
, ,
,
8 8 8
8
5
13
c. 
b.

2
d.
1
1

2
13
2 2
4
k
c.
2 4 14 16
,
,
,
3 3 3
3
sec x  csc x
1  sin x
2
 sin x  cos x
  sec x  tan x 
tan x  cot x
1  sin x
1
1
1  sin x 1  sin x

·
 RHS
cos x sin x  RHS
1  sin x 1  sin x
sin x cos x
2
1  2sin x  sin x

 RHS
cos x sin x
2
8. a.
b.
1  sin x
sin x  cos x
2
1  2sin x  sin x
 RHS
cos x sin x  RHS
cos 2 x
sin 2 x  cos 2 x
1
2sin x sin 2 x
cos x sin x


 RHS
2
2
2
cos x cos x cos x
sin x  cos x cos x sin x
·
 sin x  cos x
sec 2 x  2sec x tan x  tan 2 x  (sec x  tan x) 2
cos x sin x sin 2 x  cos 2 x
______________________________________________________________________________
cot x cot y  1
cot( x  y ) 
cot x  cot y
cos( x  y )
 RHS
sin( x  y )
 cos x cos y  sin x sin y   RHS
 sin x cos y  cos x sin y 
1
sin y sin x
 RHS
1
 sin x cos y  cos x sin y 
sin y sin x
cos x cos y
1
cot x cot y  1
sin x sin y

cot x  cot y
c. cos y  cos x
sin y sin x
 cos x cos y  sin x sin y 




sin   x   sin   x 
2

2

 
 
LHS  sin   cos  x   sin x cos  
2
2
d. LHS  1 cos x  0
LHS  cos x


sin   x   cos x
2

______________________________________________________________________________
1  sin 2 x
1
 1  sec x csc x
sin 2 x
2
sin 2 x
1

 RHS
sin 2 x sin 2 x
e.
1
1
 RHS
2sin x cos x
1
1
1  csc x sec x  1  sec x csc x
2
2