Ann and Sue both verified the same trig identity.
Ann’s Work
cos 
 sec   tan 
1  sin 
1
sin 


cos  cos 
1  sin  1  sin 


cos  1  sin 
1  sin 2 

cos  (1  sin  )

(sin 2   cos 2  )  sin 2 
cos  (1  sin)
cos 2 
cos  (1  sin  )
cos 
cos 

1  sin  1  sin 

Sue’s Work
cos 
 sec   tan 
1  sin 
cos  1  sin 


1  sin  1  sin 
cos  (1  sin  )

1  sin 2 
cos  (1  sin  )

2
(sin   cos 2  )  sin 2 
cos  (1  sin  )

cos 2 
1  sin 

cos 
1
sin 


cos  cos 
sec   tan   sec   tan 
Which student(s) verification was correct?
A
D
Ann was correct.
B
Sue was correct.
Neither Ann nor Sue were correct.
C
Both Ann and Sue were correct.
Andy simplified sin( x)  tan( x) using trig identifies.
1  sec( x)
sin( x)  tan( x)
1  sec( x)
sin x  tan x
Step 1
1  sec x
sin x
Step 2
sin x 
cos x
1
1
cos x
Step 3
sin x 

sin x 
cos x 
cos
x


cos x  1  1

cos x 

sin x cos x  sin x
Step 4
cos x  1
Step 5
sin x  cos x  1
Step 6
cos x  1
sin x
He graphed y  sin( x)  tan( x) on his calculator to verify his work.
1  sec( x)
Which is a true statement concerning Andy’s work.
A
Andy simplified the expression correctly since the graph verifies his work.
B
Andy’s graph does not verify his work. He made a mistake on step 1.
C
Andy’s graph does not verify his work. He made a mistake on step 2.
D
Andy’s graph does not verify his work. He made a mistake on step 4.
Complete the reasoning for each step of the verification.
Reason
1
sec x  tan x 
sec x  tan x
1
sec x  tan x
sec x  tan x 

sec x  tan x sec x  tan x
sec x  tan x
sec x  tan x 
sec 2 x  tan 2 x
sec x  tan x
sec x  tan x 
(tan 2 x  1)  tan 2 x
sec x  tan x
sec x  tan x 
1
sec x  tan x  sec x  tan x
The graph below represents the graph of each side of the equation cos x  sin x 1 on the interval
[0, 2 ) .
Which is not a possible solution to the equation
A
3
B
3
2
C

D


2
cos x  sin x 1 over the interval  4 ,0 ?
What are all of the solutions to
A
x  2n ,
B
sin 2 x  cos x  1  0 ?
x    2n ,
C
x    2n or x 

 n
2
,
D no solution
Given the following:
12

 
, for
13
2
15

tan    , for    

8
2
Find cos     .

A

sin  
220
221
B

140
221
C
140
221
D
220
221
2
The equation tan x  cos x  2 is solved over the interval
[0, 2 )
sec x
In which step was a mistake made?
A
Step 1
B
Step 3
C
Step 4
D
No mistake was
made.
Equation
Step 1
Step 2
Step 3
tan 2 x
 cos x  2
sec x
cos x  tan 2 x  cos x  2


cos x tan 2 x  1  2


cos x sec 2 x  2
Step 4
cos x  2
or
Step 5
cos x  2
or
sec2 x  2
sec2 x  2
sec x   2
 3 5 7
x , , ,
4 4 4 4
Step 6
Step 7
The horizontal distance in feet a ball travels after it is hit can be modeled by the function
v0
represents the initial velocity and

d
v0 2 sin  2 
32
represents the angle the ball is hit from the horizontal in degrees.
If a tennis ball, hit with an initial velocity of 45 feet per second, travels a horizontal distance of 40 feet, at
approximately what angles could the ball have been hit?
I
19.603
II
39.205
III 70.397
A
I only
B II only
C I and III
D II and III
,
Michelle solved the equation

0
y  sin 2
0
y  cos 
1



6
3
2
3
2
3
3
2
1
2
2
What are the solutions to
A

  0, , 
2
sin 2x  cos x by making a table of values for y  sin 2 x and y  cos x .
B
2
3
3

2
1

2
0
0
5
6


3
2
0

3
2
1
sin 2x  cos x on the interval [0, 2 ) ? (P.2P, P.5N)

  5
,
6 2
,
6
C

3
  0, ,  ,
2
2
D

  5 3
, , ,
6 2 6 2