2
θ
θ
θ
=
Example: Find
cos
if
sin
and
is
a
quadrant
II
angle.
5
θ
θ
+
=
2
2
Solution:
Since
,
cos
sin
1
2
2
4
21
θ
θ
+
=
=
−
=
2
2
cos
1
or
cos
1
.
5
25
25
21
21
θ
=
−
=
−
Then,
cos
,
where the - sign is chosen because the cosine is negative
25
5
in quadrant II.
1
θ
θ
θ
=
−
Example: Find
cot
if
cos
and
is
in
quadrant
III.
3
θ
θ
θ
Solution: Since none of the basic identities involves just cot
and cos
, we first find sin
and
θ
2
cos
1
θ
θ
+
−
=
2
then find
cot
as
.
Now
sin
. 1
θ
sin
3
8
2
2
θ
θ
=
=
−
The negative sign is chosen because θ is in
2
So
sin
and
sin
.
9
3
quadrant III. Finally,
−
1
θ
cos
1
2
3
θ
=
=
=
=
cot
.
θ
−
sin
4
2
2
2
2
3
1
θ
θ
θ
=
−
Example: Find
sin
if
tan
and
is
in
quadrant
IV.
2
1
θ
=
=
−
Solution: First,
cot
. 2
−
1
2
θ
θ
+
=
2
2
Second, from the Pythagorean identity
cot
1
csc
,
we get
( )
θ
−
+
=
2
2
2
1
csc
θ
=
2
csc
. 5
θ
Since θ is a quadrant IV angle,
=
−
csc
. 5
1
1
5
θ
=
=
=
−
Finally,
sin
.
θ
−
csc
5
5
Problems:
Use only the reciprocal identities, quotient identities and Pythagorean identities to solve Problems
106 – 113.
θ
θ
θ
=
−
106. Find
sec
if
tan
4
and
is
a
quadrant
IV
angle.
2
θ
θ
=
107. Find
csc
if
sin
.
7
θ
θ
=
−
108. Find
cos
if
sec
. 1
3
θ
θ
θ
=
109. Find
cot
if
csc
and
is
a
quadrant
I
angle.
2
2
θ
θ
θ
=
−
110. Find
tan
if
cos
and
is
a
quadrant
II
angle.
3
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