Download Chapter 5 Trigonometric Functions of Angles

Chapter 5 Trigonometric Functions of
Angles
Section 4 The Other Trigonometric
Functions
Setting the Stage
We will consider the relationship between the sides of a
triangle here in terms of side lengths instead of the value sine
and cosine.
Setting the Stage
We will consider the relationship between the sides of a
triangle here in terms of side lengths instead of the value sine
and cosine.
(x, y)
•
r
y
θ
x
Setting the Stage
We will consider the relationship between the sides of a
triangle here in terms of side lengths instead of the value sine
and cosine.
(x, y)
•
r
y
θ
x
sin(θ ) =
Setting the Stage
We will consider the relationship between the sides of a
triangle here in terms of side lengths instead of the value sine
and cosine.
(x, y)
•
r
y
θ
x
sin(θ ) =
y
r
Setting the Stage
We will consider the relationship between the sides of a
triangle here in terms of side lengths instead of the value sine
and cosine.
(x, y)
•
r
y
θ
x
sin(θ ) =
cos(θ ) =
y
r
Setting the Stage
We will consider the relationship between the sides of a
triangle here in terms of side lengths instead of the value sine
and cosine.
(x, y)
•
r
y
θ
x
sin(θ ) =
cos(θ ) =
y
r
x
r
Other Trig Functions
(x, y)
•
r
y
θ
x
Cosecant
csc(θ ) =
Other Trig Functions
(x, y)
•
r
y
θ
x
Cosecant
csc(θ ) =
1
=
sin(θ )
Other Trig Functions
(x, y)
•
r
y
θ
x
Cosecant
csc(θ ) =
1
r
=
sin(θ )
y
Other Trig Functions
(x, y)
•
r
y
θ
x
Secant
sec(θ ) =
Other Trig Functions
(x, y)
•
r
y
θ
x
Secant
sec(θ ) =
1
=
cos(θ )
Other Trig Functions
(x, y)
•
r
y
θ
x
Secant
sec(θ ) =
1
r
=
cos(θ )
x
Other Trig Functions
(x, y)
•
r
y
θ
x
Tangent
tan(θ ) =
Other Trig Functions
(x, y)
•
r
y
θ
x
Tangent
tan(θ ) =
sin(θ )
=
cos(θ )
Other Trig Functions
(x, y)
•
r
y
θ
x
Tangent
tan(θ ) =
sin(θ )
y
=
cos(θ )
x
Other Trig Functions
(x, y)
•
r
y
θ
x
Cotangent
cot(θ ) =
Other Trig Functions
(x, y)
•
r
y
θ
x
Cotangent
cot(θ ) =
cos(θ )
=
sin(θ )
Other Trig Functions
(x, y)
•
r
y
θ
x
Cotangent
cot(θ ) =
cos(θ )
x
=
sin(θ )
y
Examples
Example
Evaluate tan(225◦ ).
Examples
Example
Evaluate tan(225◦ ).
The reference angle is
Examples
Example
Evaluate tan(225◦ ).
The reference angle is 45◦ .
Examples
Example
Evaluate tan(225◦ ).
The reference angle is 45◦ .
In what quadrant is 225◦ ?
Examples
Example
Evaluate tan(225◦ ).
The reference angle is 45◦ .
In what quadrant is 225◦ ? quadrant III
Examples
Example
Evaluate tan(225◦ ).
The reference angle is 45◦ .
In what quadrant is 225◦ ? quadrant III
tan(45◦ ) =
Examples
Example
Evaluate tan(225◦ ).
The reference angle is 45◦ .
In what quadrant is 225◦ ? quadrant III
tan(45◦ ) =
sin(45◦ )
cos(45◦ )
=
Examples
Example
Evaluate tan(225◦ ).
The reference angle is 45◦ .
In what quadrant is 225◦ ? quadrant III
tan(45◦ ) =
sin(45◦ )
cos(45◦ )
√
=
2
√2
2
2
=
Examples
Example
Evaluate tan(225◦ ).
The reference angle is 45◦ .
In what quadrant is 225◦ ? quadrant III
tan(45◦ ) =
sin(45◦ )
cos(45◦ )
√
=
2
√2
2
2
=1
Examples
Example
Evaluate tan(225◦ ).
The reference angle is 45◦ .
In what quadrant is 225◦ ? quadrant III
tan(45◦ ) =
sin(45◦ )
cos(45◦ )
√
=
2
√2
2
2
=1
So, tan(225◦ ) =
Examples
Example
Evaluate tan(225◦ ).
The reference angle is 45◦ .
In what quadrant is 225◦ ? quadrant III
tan(45◦ ) =
sin(45◦ )
cos(45◦ )
√
=
2
√2
2
2
=1
So, tan(225◦ ) = 1
Examples
Example
Evaluate csc
5π
6
Examples
Example
Evaluate csc
5π
6
The reference angle is
Examples
Example
Evaluate csc
5π
6
The reference angle is
π
6.
Examples
Example
Evaluate csc
5π
6
The reference angle is
In what quadrant is
π
6.
5π
6 ?
Examples
Example
Evaluate csc
5π
6
The reference angle is
In what quadrant is
π
6.
5π
6 ?
quadrant II
Examples
Example
Evaluate csc
5π
6
The reference angle is
In what quadrant is
π
6.
5π
6 ?
csc
quadrant II
π
6
=
Examples
Example
Evaluate csc
5π
6
The reference angle is
In what quadrant is
π
6.
5π
6 ?
csc
quadrant II
π
6
=
1
sin
π
6
=
Examples
Example
Evaluate csc
5π
6
The reference angle is
In what quadrant is
π
6.
5π
6 ?
csc
quadrant II
π
6
=
1
sin
π
6
=
=
1
1
2
Examples
Example
Evaluate csc
5π
6
The reference angle is
In what quadrant is
π
6.
5π
6 ?
csc
quadrant II
π
6
=
1
sin
π
6
=
1
1
2
=2
Examples
Example
Evaluate csc
5π
6
The reference angle is
In what quadrant is
π
6.
5π
6 ?
csc
quadrant II
π
6
=
1
sin
π
6
=
1
1
2
=2
So, csc
5π
6
=
Examples
Example
Evaluate csc
5π
6
The reference angle is
In what quadrant is
π
6.
5π
6 ?
csc
quadrant II
π
6
=
1
sin
π
6
=
1
1
2
=2
So, csc
5π
6
=2
Examples
Example
Evaluate cot
11π
3
Examples
Example
Evaluate cot
11π
3
The reference angle is
Examples
Example
Evaluate cot
11π
3
The reference angle is
π
3.
Examples
Example
Evaluate cot
11π
3
The reference angle is
In what quadrant is
π
3.
11π
3 ?
Examples
Example
Evaluate cot
11π
3
The reference angle is
In what quadrant is
π
3.
11π 5π
3 ? 3 ?
Examples
Example
Evaluate cot
11π
3
The reference angle is
In what quadrant is
π
3.
11π 5π
3 ? 3 ?
quadrant IV
Examples
Example
Evaluate cot
11π
3
The reference angle is
In what quadrant is
π
3.
11π 5π
3 ? 3 ?
cot
π
3
quadrant IV
=
Examples
Example
Evaluate cot
11π
3
The reference angle is
In what quadrant is
π
3.
11π 5π
3 ? 3 ?
cot
π
3
quadrant IV
cos
=
sin
=
π
3
π
3
Examples
Example
Evaluate cot
11π
3
The reference angle is
In what quadrant is
π
3.
11π 5π
3 ? 3 ?
cot
π
3
quadrant IV
cos
=
sin
=
=
π
3
π
3
1
√2
3
2
Examples
Example
Evaluate cot
11π
3
The reference angle is
In what quadrant is
π
3.
11π 5π
3 ? 3 ?
cot
π
3
quadrant IV
cos
=
sin
=
π
3
π
3
1
√2
3
2
1
=√
3
Examples
Example
Evaluate cot
11π
3
The reference angle is
In what quadrant is
π
3.
11π 5π
3 ? 3 ?
cot
π
3
quadrant IV
cos
=
sin
=
π
3
π
3
1
√2
3
2
1
=√
3
So, cot
11π
3
=
Examples
Example
Evaluate cot
11π
3
The reference angle is
In what quadrant is
π
3.
11π 5π
3 ? 3 ?
cot
π
3
quadrant IV
cos
=
sin
=
π
3
π
3
1
√2
3
2
1
=√
3
So, cot
11π
3
= − √13
Identities
We all remember the Pythagorean Identity, right?
Identities
We all remember the Pythagorean Identity, right?
The Pythagorean Identity
sin2 θ + cos2 θ = 1
Identities
We all remember the Pythagorean Identity, right?
The Pythagorean Identity
sin2 θ + cos2 θ = 1
We can use variations of this to prove different equations hold.
Identities
We all remember the Pythagorean Identity, right?
The Pythagorean Identity
sin2 θ + cos2 θ = 1
We can use variations of this to prove different equations hold.
Variation 1
1 + cot2 θ = csc2 θ
Identities
We all remember the Pythagorean Identity, right?
The Pythagorean Identity
sin2 θ + cos2 θ = 1
We can use variations of this to prove different equations hold.
Variation 1
1 + cot2 θ = csc2 θ
Variation 2
tan2 θ + 1 = sec2 θ
Identities
Example
Simplify secθcosθ − cos2 θ.
Identities
Example
Simplify secθcosθ − cos2 θ.
secθcosθ − cos2 θ
Identities
Example
Simplify secθcosθ − cos2 θ.
secθcosθ − cos2 θ =
1
cosθ − cos2 θ
cosθ
Identities
Example
Simplify secθcosθ − cos2 θ.
secθcosθ − cos2 θ =
1
cosθ − cos2 θ
cosθ
= 1 − cos2 θ
Identities
Example
Simplify secθcosθ − cos2 θ.
secθcosθ − cos2 θ =
1
cosθ − cos2 θ
cosθ
= 1 − cos2 θ
= sin2 θ
Identities
Example
Show that cos2 θ + cos2 θtan2 θ = 1.
Identities
Example
Show that cos2 θ + cos2 θtan2 θ = 1.
cos2 θ + cos2 θtan2 θ
Identities
Example
Show that cos2 θ + cos2 θtan2 θ = 1.
cos2 θ + cos2 θtan2 θ = cos2 θ + cos2 θ
sin2 θ
cos2 θ
Identities
Example
Show that cos2 θ + cos2 θtan2 θ = 1.
sin2 θ
cos2 θ
2
= cos θ + sin2 θ
cos2 θ + cos2 θtan2 θ = cos2 θ + cos2 θ
Identities
Example
Show that cos2 θ + cos2 θtan2 θ = 1.
sin2 θ
cos2 θ
2
= cos θ + sin2 θ
cos2 θ + cos2 θtan2 θ = cos2 θ + cos2 θ
=1
Identities
Example
Simplify sin4 θ − cos4 θ.
Identities
Example
Simplify sin4 θ − cos4 θ.
sin4 θ − cos4 θ
Identities
Example
Simplify sin4 θ − cos4 θ.
sin4 θ − cos4 θ = (sin2 θ + cos2 θ )(sin2 θ − cos2 θ )
Identities
Example
Simplify sin4 θ − cos4 θ.
sin4 θ − cos4 θ = (sin2 θ + cos2 θ )(sin2 θ − cos2 θ )
= 1(sin2 θ − cos2 θ )
=
Identities
Example
Simplify sin4 θ − cos4 θ.
sin4 θ − cos4 θ = (sin2 θ + cos2 θ )(sin2 θ − cos2 θ )
= 1(sin2 θ − cos2 θ )
= sin2 θ − (1 − sin2 θ )
Identities
Example
Simplify sin4 θ − cos4 θ.
sin4 θ − cos4 θ = (sin2 θ + cos2 θ )(sin2 θ − cos2 θ )
= 1(sin2 θ − cos2 θ )
= sin2 θ − (1 − sin2 θ )
= 2sin2 θ − 1
Identities
Example
Simplify
1−cos2 θ
.
sec2 θ −1
Identities
Example
Simplify
1−cos2 θ
.
sec2 θ −1
Whenever we see the square of a trig function ±1, we want to
replace it with another square of a trig function when possible.
Identities
Example
Simplify
1−cos2 θ
.
sec2 θ −1
Whenever we see the square of a trig function ±1, we want to
replace it with another square of a trig function when possible.
1 − cos2 θ
sec2 θ − 1
Identities
Example
Simplify
1−cos2 θ
.
sec2 θ −1
Whenever we see the square of a trig function ±1, we want to
replace it with another square of a trig function when possible.
sin2 θ
1 − cos2 θ
=
sec2 θ − 1
tan2 θ
Identities
Example
Simplify
1−cos2 θ
.
sec2 θ −1
Whenever we see the square of a trig function ±1, we want to
replace it with another square of a trig function when possible.
sin2 θ
1 − cos2 θ
=
sec2 θ − 1
tan2 θ
cos2 θ
= sin2 θ ·
sin2 θ
Identities
Example
Simplify
1−cos2 θ
.
sec2 θ −1
Whenever we see the square of a trig function ±1, we want to
replace it with another square of a trig function when possible.
sin2 θ
1 − cos2 θ
=
sec2 θ − 1
tan2 θ
cos2 θ
= sin2 θ ·
sin2 θ
= cos2 θ