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Trigonometric Values of an Angle in
Standard Position
120º 90º
135º
Quadrant II
60º
Quadrant I
45º
150º
30º
180º
0º
210º
Quadrant III
330º
225º
240º
radius = r
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315º
270º 300º
Quadrant IV
An angle , in standard position is shown below. Let P
(x, y) be any point on the terminal arm of any angle , in
y
standard position.
y
sin θ 
r
x
cos θ 
r
y
tan θ 
x
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P( x, y )
r
y

x
r  x2  y 2
x
Example 1: Point P(4, 3) lies
on the terminal arm of angle .
Determine the sin, cos
and tan of angle  and
the measure of the
principal angle.
x = 4, y = 3
r  4 5
2
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P( 4, 3)
5
3

2
r=5
y
sin θ 
r
3

5
y
x
y
cos θ 
tan θ 
r
x
3
4


4
5
0
4
3
  sin  
5
= 37º
1
x
Example 2: Point P(– 4, 3) lies
on the terminal arm of angle .
Determine the sin, cos
and tan of angle  and
P(– 4, 3)
the measure of the
principal angle.
3
x = 4, y = 3
y
sin θ 
r
3

5
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r  (4) 2  (3) 2
r=5
x
y
cos θ 
tan θ 
r
x
3
4


4
5
y
5
–4

0
3
  sin  
5
= 37º
 = 180 – 37º
1
x
Example 3: Point P(– 4, – 3) lies on
the terminal arm of angle .
Determine the sin, cos
and tan of angle  and
the measure of the
principal angle.
–4
x = – 4, y = – 3 r  (4)2  (3) 2
r=5
y

0
–3
5
x
y
y
cos θ 
sin θ 
tan θ 
r
r
x P(– 4, – 3)
1  3 
  sin  
3
3
4



5
5
4
= 37º
5
= 180 + 37º = 217º
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x
Example 4: Point P( 4, –3) lies
on the terminal arm of angle .
Determine the sin, cos
and tan of angle  and
the measure of the
principal angle.
x = – 4, y = 3 r  (4)2  (3) 2
r=5
x
y
cos θ 
tan θ 
r
x
3
3
4



5
4
5
y
sin θ 
r
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y
4
x

5
–3
0
P(– 4, –3)
3
  sin  
5
= 37º
 = 360º – 37º = 323º
1
Summarize what you have learned in the table below.
Quadrant Sign of Sign of Sign of Sign of Sign of
x
y
for 
sin 
cos 
tan 
I
II
III
IV
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Summarize what you have learned in the table below.
Quadrant Sign of Sign of Sign of Sign of Sign of
x
y
for 
sin 
cos 
tan 
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I
+
+
+
+
+
II
–
+
+
–
–
III
–
–
–
–
+
IV
+
–
–
+
–
Example 5:
1
Determine the value of  if sin  =
2
0 <  < 360º
y
sin  = 0.5
1 = 30º
2 = 180º – 30º
2 = 150º
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2
1
2
1
x
Example 6: Point P(– 6, –2) lies
on the terminal arm of angle .
Determine the sin, cos
and tan of angle  and
the measure of the
principal angle.
x = – 6, y = –2 r  (6)2  (2)2
r  40
–2
y

–6
0
40
x
y
cos θ 
tan θ 
r
x P(–6,–2)
1  1 
  tan  
2
1
6



3
3
40
40
= 18º
y
sin θ 
r
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 = 180 + 18º
= 198º
x
Positive Values
Sine positive
120º 90º
135º
(180 - )
All positive
60º

45º
150º
30º
180º
0º 360º
(180 + ) 210º
330º
225º
Tan positive
240º
315º
270º 300º
CAST Rule
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(360 - )
Cosine positive