Download TEST 5_1 -5_3 Take home practice KEY

Course 4
Pre-Calculus
Name:____________________________
CHAPTER 5 TAKE HOME TEST PRACTICE
Date:_______________ Period:_______
A
COMPLETE
EACH PROBLEM IN YOUR NOTES.
1. Use the given values to evaluate (if possible) all six trigonometric functions of the angle.
tan  
a)
6
5
cos  0
b)
6 61
5 61
cos   
61
61
61
61
5
csc   
sec   
cot  
6
5
6
csc(

2
 x)  3
cos  
sin   
csc  
3 2
4
2 2
3
sin  
1
3
tan   2 2
sec   3 cot  
2
4
2. Use the fundamental identities to simplify the following expressions.
a.
sin3   sin  cos2   sin 
b.
sin  (sin 2   cos 2  )  sin  (1) 
sin 

d. tan   x  sec x
2

cot x(sec x) 
sec2 x  1
sec x  1
(sec x  1)(sec x  1)
 sec x  1
(sec x  1)
e.
1
cot 2 x  1
f. csc 4   2 cot 2   4
2
cos x  1 


sin x  cos x 
1
 sin 2 x
2
csc x
csc 4   2(csc 2   1)  4 
csc 4   2 csc 2   2  4 
csc 4   2 csc 2   2
3. Verify the following trigonometric identities:
or
csc  csc   2 cot 2   4 
2
sec x cot x  cot x  tan x
2
cos x sin 2 x
cot x(sec x  1)  cot x(tan x) 


sin x cos 2 x
sin x
 tan x
cos x
2
2


 x 1
2

2
2
b. sin x  sin 
sin 2 x  cos 2 x  1 pythgorean identity
sin 2   cos2 
= 1 + cot α
sin 2   sin  cos 
2
(1  cot 2  )(1  cot 2  )  2 cot 2   4 
1  2 cot 2   cot 4   2 cot 2   4  cot 4   3
c.
d.
sec 2 x  tan 2 x  1
cot


2
csc   1
1
 csc x
sin x
a.
sec 4 x  tan 4 x
sec 2 x  tan 2 x
(sec 2 x  tan 2 x)(sec 2 x  tan 2 x)

(sec 2 x  tan 2 x)
c.
cos 
1  sin 

 2 sec 
1  sin 
cos 
cos 2   (1  2 sin   sin 2  ) (cos 2   sin 2  )  1  2 sin 


cos  (1  sin  )
cos  (1  sin  )
2  2 sin 
2(1  sin  )
2


 2 sec 
cos  (1  sin  ) cos  (1  sin  ) cos 
(sin   cos  )(sin   cos  ) sin  cos 


 1  cot 
sin  (sin   cos  )
sin  sin 
4. Rewrite sin 4 x tan 2 x in terms of cosine function only.
 sin 2 x  sin 2 x sin 2 x sin 2 x (1  cos 2 x)(1  cos 2 x)(1  cos 2 x) (1  cos 2 x) 3
 
sin 4 x




2
1
1
cos 2 x
cos 2 x
cos 2 x
 cos x 
5. For each equation, find solutions in the interval [0, 2 ) and all general solutions.
b. 3 3 tan x  3
a. csc 2 x  csc x  2  0
sin x  tan x
(csc x  2)(csc x  1)  0
csc x  2
csc x  1
1
2
 5
x ,
6 6
sin x  1
sin x 
x
3
3
tan x 


3
3 3
3
3
2
x
2
cos x  1
6
x0
,
6
f. sec 2 x  sec x  2
sec 2 x  sec x  2  0
1
sin x 
2
cot x(cos 2 x  2)  0
(sec x  2)(sec x  1)  0
sec x  2
cos x  2
tan x  undefined
 3
,
2 2
2
cos x   2
x
 does not exist
g. sec 2 x  tan x  3
 5
,
6 6
csc 2 x 
tan x tan x  2  0
2
(tan x  2)(tan x  1)  0
csc x  
tan x  1
this angle is not formed
by a special right triangle
arctan( 2)  1.1radians
x
or  1.1    2.03radians
i. cos 4x(cos x – 1)=0
1
2
2 4
x
,
3 3
cos x  
sec x  1
cos x  1
x0
h. 3 csc 2 x  4  0
(1  tan 2 x)  tan x  3  0
tan x  2
sin x sin x

1
cos x
sin x cos x  sin x
 7
cot x cos 2 x  2 cot x  0
x
1
e. 2sin x  1  0
d. cot x cos x  2cot x
cot x  0
c. sin x  tan x  0
 5
4
,
4
4
3
2
do not rationaliz e this value!
3
you need reciprocal of sin! !!
3
make sure you include  when taki ng
2
 2 4 5
x ,
,
,
3 3 3 3
sin x  
(cos 4 x)(cos x  1)  0 already factored! for us!
cos( 4 x)  0
cos x  1

3
cos( )  0 & cos( )
2
2
therefore
4x 
x

& 4x 
2
 3
8
,
8
3
solve for x
2
x0