Precalculus - Unit 5 Exam – 60 Points
Name _________________________________
Verify the following Identities. (5 points each)
cot x
= cos x
csc x
" cos x %
$
'
# sin x &
...
1) " 1 %
$
'
# sin x &
" cos x %" sin x %
...$
'$
'
# sin x &# 1 &
...cos x
sin x cot x sec x = 1
sin x cos x
1
"
"
2) ...
1 sin x cos x
...1
!
sin 2 x cot 4 x tan 2 x = 1" sin 2 x
!
2
3)
4
(sec x + tan x)(sec x " tan x ) = 1
2
sin x cos x sin x
#
#
1
sin 4 x cos2 x
...cos2 x
...
2
2
4) ...sec x " tan x
...1+ tan 2 x " tan 2 x
...1
...1" sin 2 x
!
!
!
csc x " cot x cos x = sin x
1
cos x cos x
...
"
#
sin x sin x
1
2
1" cos x
5) ...
sin x
sin 2 x
...
sin x
...sin x
2 " 5cot x 2sin x " 5cos x
=
2 + 5cot x 2sin x + 5cos x
cos x
2"5
sin x
...
cos x
2+5
sin x
7)
#
cos x &
2"5
%
(
sin x
sin x (
...
%
cos x (
sin x %
2+5
$
sin x '
2sin x " 5cos x
...
2sin x + 5cos x
!
!
!
cos x + tan x
= cot x + sec x
sin x
cos x tan x
...
+
sin x sin x
sin x
6) ...cot x + cos x
sin x
sin x
1
...cot x +
"
cos x sin x
...cot x + sec x
sin x
sin x
"
= 2cot x
1" cos x 1+ cos x
# 1+ cos x &# sin x & # sin x &#1" cos x &
...%
(%
( "%
(%
(
$ 1+ cos x '$ 1" cos x ' $1+ cos x '$1" cos x '
sin x + sin x cos x " sin x + sin x cos x
8) ...
1" cos 2 x
2sin x cos x
...
sin 2 x
2cos x
...
sin x
...2cot x
9) Find the exact value of (5 points)
sin (300° + 45°)
cos345°
cos300°cos45° " sin 300°sin 45°
sin 300°cos45° + cos300°sin 45°
b. 1 # 2 " " 3 # 2
2 2
2
2
2+ 6
4
a. " 3 # 2 + 1 # 2
2
2 2 2
2" 6
4
tan 345°
c)
2" 6
4
2+ 6
4
2" 6
2+ 6
!
3
12
10) Given cos A = " , A in quadrant III and sin B = , B in quadrant II, find
5
13
!
a. sin ( A " B)
b. cos( A " B)
c. tan( A " B)
!
56
sin Acos B " cos Asin B cos Acos B + sin Asin B
65
# "4 &# "5 & # 3 &#12 &
# "3 &# "5 & # "4 &# 12 &
"33
% (% ( " % " (% (
% (% ( + % (% (
65
$ 5 '$ 13 ' $ 5 '$ 13 '
$ 5 '$ 13 ' $ 5 '$ 13 '
"56
56
33
=
"
33
65
65
!
6
11) Given sin A = , A in quadrant II find
! 7
a. sin 2 A
b. cos2 A
!
So cos
A
9
="
2
34
or
"3
34
A
< 135°
2
d. quadrant of ( A " B)
ASTC...Quadrant II
(6 pts)
c. tan 2 A
"12 13
cos2 A " sin 2 A
2sin Acos A
49 = "12 13 # "49
# " 13 &2 # 6 &2
" 6 %" ( 13 %
"23
49
23
%%
(( " % (
''
2$ '$$
49
# 7 &# 7 &
$ 7 ' $ 7'
12 13
"23
(12 13
23
49
49
!
"15
A
12) Given sin A =
, A in quadrant III find cos
(3 pts)
17
2
!
!
# "8 &
%1+ ( #17 &
A
1+ cos A
9
cos = ±
= ± % 17 ( % ( = ±
2
2
34
% 2 ( $17 '
$
'
Since 180° < A < 270°, 90° <
(6 pts)
d. quadrant of 2A
ASTC...Quadrant III