Name: ____________________ (
)
Class:
Date:
Lesson Worksheet 6.1B(I)
Objective: To determine the signs of trigonometric ratios.
The signs of trigonometric ratios can be determined by the ‘CAST’ diagram.
y
All
are positive
Sin
is positive
Quadrant II
Tan
is positive
O
Quadrant III
1.
x
Cos
is positive
Quadrant IV
In each of the following, determine all the possible quadrants in which the angle θ lies.
(a) sin θ =
1
3
(b) sin θ = –
>
∵ sin θ ________
0 (< / >)
∴ θ lies in quadrant
I
or
II
.
∴ θ lies in quadrant
∴ θ lies in quadrant
or
IV
.
I
or
IV
.
∵ tan θ ________
0 (< / >)
<
or
IV
.
∴ θ lies in quadrant
II
4
and 90º < θ < 180º, find the values of cos θ and tan θ.
5
If sin θ =
<θ<
90º
y
180º
∴ θ lies in quadrant
P(a, b)
.
II
Let P(a, b) be a point on the terminal side of θ.
Let b =
III
(d) tan θ = –2
∵ cos θ ________
0 (< / >)
>
∵
1
2
∵ sin θ ________
0 (< / >)
<
(c) cos θ = 0.4
2.
Quadrant I
and r =
4
5
.
5
r=
a 2  b2
=
a 2  42
( 5 )2 = a2 +
a2 =
5
42
9
θ
0
x
5
4
a
a = 3 (rejected) or –3
44
cos θ =
3
a
= 5
r
tan θ =
4
b
= 
3
a
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Chapter 6
3.
1
and 270º < θ < 360º, find the values of sin θ and tan θ.
4
(Leave the answers in surd form.)
If cos θ =
∵
<θ<
270º
y
360º
∴ θ lies in quadrant
.
IV
θ
Let P(a, b) be a point on the terminal side of θ.
Let a =
1
and r =
a b
=
4
.
4
2
r=
4
2
1
P(a, b)
 b2
12
( 4 )2 = ( 1 )2 + b2
b2 =
x
0
4
b
15
b = 15 (rejected) or – 15
b  15
15
sin θ = =
=
r
4
4
tan θ =
4.
b  15
=
=  15
a
1
If tan θ = 3 and 180º < θ < 270º, find the values of sin θ and cos θ.
Exercise 6.1: 10, 11
(Leave the answers in surd form.)
∵
<θ<
180º
270º
∴ θ lies in quadrant
III
y
.
Let P(a, b) be a point on the terminal side of θ.
Let a =
–1
and b =
r = a b
2
–3
θ
.
0
2
x
= (1) 2  (3) 2
P(a, b)
= 10
3
3
b
sin θ = =
=
r
10
10
cos θ =
a
1
1
=
=
r
10
10
Try More
5.
If cos θ = 
2
13
and 90º < θ < 180º, find the values of sin θ and tan θ.
(Leave the answers in surd form if necessary.)
∵ 90º < θ < 180º
∴ θ lies in quadrant II.
Let P(a, b) be a point on the terminal side of θ.
Let a = –2 and r = 13 .
sin θ =
3
b
=
r
13
tan θ =
3
b
3
=
=
a 2
2
r = a 2  b2
13 = ( 2) 2  b 2
( 13 ) 2 = (–2)2 + b2
b2 = 9
b = 3 or –3(rejected)
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45