Section 6.3 properties of the trigonometric functions

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Section 6.3
Properties of the
Trigonometric Functions

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If sin θ > 0 and cos θ < 0, name the quadrant in which the
angle θ lies.
For sin θ > 0 the y value must be positive so the angle must
be in quadrant I or II.
For cos θ < 0 the x value must be negative so the angle must
be in quadrant II or III.
Therefore, this angle must lie in quadrant II.

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10 3 10
Given sin and cos , find the value of each of the four remaining
10 10
trigonometric functions of .
θ θ
θ
= =
10
sin 10 10 110tan
cos 10 33 10 3 10
10
θ
θ
θ
= = = × =
1 1
cot 3
1tan
3
θ
θ
= = =
1 1 10
csc 10
sin 1 1
10
0 0
θ
θ
= = = =
1 1 10 10
sec
cos 33 10 3 10
10
θ
θ
= = = =

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2
2
Find the exact value of each expression. Do not use a calculator.
cos
1 3(a) cos 35 (b) cot
csc 35 3sin
3
π
π
π
+ ° −
°
2 2 2
2
1
(a) cos 35 sin 35 cos 35 1
csc 35
+ ° = °+ ° =
°
cos
3(b) cot cot cot 0
3 3 3sin
3
π
π π π
π
− = − =

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2
Given that sin and cos , find the exact value of each
5
of the remaining five trigonometric functions of .
θ θ < 0
θ
=
21
cos =
5
x
r
θ
−
=
5
csc =
2
r
y
θ =
5 5 21
sec =
2121
r
x
θ = = −
−
21
cot =
2
x
y
θ = −
θ
P(x,2)
r = 5
2 2 21
tan =
2121
y
x
θ = = −
−
2
Since sin 0 and cos 0, is in quadrant II.
5
θ θ θ= > <
2 2 2
The circle has a radius of 5 so its equation is 5x y+ =
2 2 2
is a point on the circle so 5 2P x= +
21x = −2
21x =

25. 2
Given that sin and is an acute angle, find the exact value of each
5
of the remaining five trigonometric functions of .
θ θ
θ
=
2 2
sin c 1osθ θ+ =
2
22
co 1s
5
θ
 
+ = ÷
 
2 4 21
1
2 2
cos
5 5
θ − ==
21 21
2 5
cos
5
θ == −−
Since θ is in quadrant II, x
values are negative
2
sin 2 55tan
cos 521 21
5
2 21
21
θ
θ
θ
 
= = = × − = − ÷
 
−
1 21
cot
tan 2
θ
θ
= = −
1 1 5
csc
2sin 2
5
θ
θ
= = =
1 1 5 5 21
sec
cos 2121 21
5
θ
θ
= = = − = −
−

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27. P(-1,-3)
θ
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1
Given that cot and sin , find the exact value of each
3
of the remaining five trigonometric functions of .
θ θ < 0
θ
=
1 10
cos =
1010
x
r
θ
−
= = −
3
tan = 3
1
y
x
θ
−
= =
−
10
csc =
3
r
y
θ = −
10
sec = 10
1
r
x
θ = = −
−
3 3 10
sin =
1010
y
r
θ
−
= = −
1
Since cot 0 and sin 0, is in quadrant III.
3
θ θ θ= > <
2
is a point on the circle so 1 9 10P r = + =
10r =

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2 2
cot 1 cscθ θ+ =
2
21
1 csc
3
θ
 
+ = ÷
 
2 1
csc
9
10
1
9
θ = + =
10 10
9 3
cscθ == −−
Since sin θ < 0, csc θ is negative.
sin 3 10 110cos
tan 3
3 10
10
110 3 0
θ
θ
θ
−
= = = − × = −
1 1
tan 3
1cot
3
θ
θ
= = =
1 1 3 3 10
sin
csc 1010 10
3
θ
θ
= = = − = −
−
1 1 10
sec 10
cos 10 1
10
0
θ
θ
= = = − = −
−
1
Given that cot and sin , find the exact value of each
3
of the remaining five trigonometric functions of .
θ θ < 0
θ
=

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Find the exact value of:
(a) cos (−60°) (b) sin (−390°) (c) tan
37
4
π 
− ÷
 
( )
1
(a) cos 60 cos60
2
− ° = ° =
( ) ( )
3
(b) sin 390 sin390 sin 30 360 sin30
2
− ° = − ° = − °+ ° = − ° = −
37 37 36
(c) tan tan tan
4 4 4 4
π π π π     
− = − = − + ÷  ÷  ÷
     
tan 9 tan 1
4 4
π π
π
 
= − + = − = − ÷
 