Download Finding trigonometric ratios for certain angles

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Finding trigonometric ratios for certain angles
(mα+hs)Smart Workshop
Semester 2, 2016
Geoff Coates
These slides describe some quick ways to:
find trigonometric ratios for the angles
π π
, 4
6
and
π
,
3
use these to find trig ratios for related angles greater than
π
2
and
solve trigonometric equations.
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Contents
Trigonometric ratios for certain angles
Trigonometric ratios for angles >
Go
π
2
Go
Go
The unit circle
Solving trigonometric equations
Go
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Trigonometric ratios for certain angles
The trigonometric ratios (sine, cosine and tangent) for angles are usually infinite decimals
but some have exact values.
Finding
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trigonometric
Workshop
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Semester
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2, 2016
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Trigonometric ratios for certain angles
The trigonometric ratios (sine, cosine and tangent) for angles are usually infinite decimals
but some have exact values. For example,
sin
π
3
√
=
3
2
exactly (rather than 0.866 . . .).
The other key angles whose trig ratios are exact are
Finding
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π
4
Contents
and
π
.
6
Prev
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5 / 11
Trigonometric ratios for certain angles
The trigonometric ratios (sine, cosine and tangent) for angles are usually infinite decimals
but some have exact values. For example,
sin
π
3
√
=
3
2
exactly (rather than 0.866 . . .).
The other key angles whose trig ratios are exact are
π
4
and
π
.
6
It’s handy to know these exact values but memorizing stuff you don’t understand is
difficult (and dull).
Finding
((mα+hs)Smart
trigonometric
Workshop
ratios
Semester
for certain
2, 2016
angles
)
Contents
Prev
Next
5 / 11
Trigonometric ratios for certain angles
The trigonometric ratios (sine, cosine and tangent) for angles are usually infinite decimals
but some have exact values. For example,
sin
π
3
√
=
3
2
exactly (rather than 0.866 . . .).
The other key angles whose trig ratios are exact are
π
4
and
π
.
6
It’s handy to know these exact values but memorizing stuff you don’t understand is
difficult (and dull). However, being able to quickly work out stuff using knowledge you
already have is easy (and fun).
Finding
((mα+hs)Smart
trigonometric
Workshop
ratios
Semester
for certain
2, 2016
angles
)
Contents
Prev
Next
5 / 11
Trigonometric ratios for certain angles
The trigonometric ratios (sine, cosine and tangent) for angles are usually infinite decimals
but some have exact values. For example,
sin
π
3
√
=
3
2
exactly (rather than 0.866 . . .).
The other key angles whose trig ratios are exact are
π
4
and
π
.
6
It’s handy to know these exact values but memorizing stuff you don’t understand is
difficult (and dull). However, being able to quickly work out stuff using knowledge you
already have is easy (and fun).
The following two right angle triangles (and some very basic knowledge about the
definition of sine, cosine and tangent) will make your life a lot easier.
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Trigonometric ratios for certain angles
2
2
1
1
2
Start with a square of side length 1 and an equilateral triangle of side length 2.
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Trigonometric ratios for certain angles
2
1
1
1
Start with a square of side length 1 and an equilateral triangle of side length 2.
Cut both in half as shown to create right-angle triangles.
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Trigonometric ratios for certain angles
2
h
d
1
1
1
Start with a square of side length 1 and an equilateral triangle of side length 2.
Cut both in half as shown to create right-angle triangles.
Work out the missing side lengths using Pythagoras’ Theorem:
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Trigonometric ratios for certain angles
2
h
d
1
1
1
Start with a square of side length 1 and an equilateral triangle of side length 2.
Cut both in half as shown to create right-angle triangles.
Work out the missing side lengths using Pythagoras’ Theorem:
h 2 = 1 2 + 12
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Trigonometric ratios for certain angles
√
2
2
d
1
1
1
Start with a square of side length 1 and an equilateral triangle of side length 2.
Cut both in half as shown to create right-angle triangles.
Work out the missing side lengths using Pythagoras’ Theorem:
h 2 = 1 2 + 12
√
So h = 2
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Trigonometric ratios for certain angles
√
2
2
d
1
1
1
Start with a square of side length 1 and an equilateral triangle of side length 2.
Cut both in half as shown to create right-angle triangles.
Work out the missing side lengths using Pythagoras’ Theorem:
h 2 = 1 2 + 12
√
So h = 2
22 = 1 2 + d 2
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Trigonometric ratios for certain angles
√
2
√
2
1
1
3
1
Start with a square of side length 1 and an equilateral triangle of side length 2.
Cut both in half as shown to create right-angle triangles.
Work out the missing side lengths using Pythagoras’ Theorem:
h 2 = 1 2 + 12
√
So h = 2
22 = 1 2 + d 2
√
So d = 3
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Trigonometric ratios for certain angles
√
2
√
2
1
1
3
1
Start with a square of side length 1 and an equilateral triangle of side length 2.
Cut both in half as shown to create right-angle triangles.
Work out the missing side lengths using Pythagoras’ Theorem:
h 2 = 1 2 + 12
√
So h = 2
22 = 1 2 + d 2
√
So d = 3
It should be clear what the angles are in both triangles.
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Trigonometric ratios for certain angles
√
2
π
4
π
6
2
1
1
√
3
π
3
1
Start with a square of side length 1 and an equilateral triangle of side length 2.
Cut both in half as shown to create right-angle triangles.
Work out the missing side lengths using Pythagoras’ Theorem:
h 2 = 1 2 + 12
√
So h = 2
22 = 1 2 + d 2
√
So d = 3
It should be clear what the angles are in both triangles.
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Trigonometric ratios for certain angles
√
2
π
4
π
6
2
1
√
3
π
3
1
1
We know that, for right angle triangles,
sin(θ) =
opposite
hypotenuse
cos(θ) =
adjacent
hypotenuse
tan(θ) =
opposite
adjacent
so we can easily find the ratios for these three angles:
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Trigonometric ratios for certain angles
√
2
π
4
π
6
2
1
√
3
π
3
1
1
We know that, for right angle triangles,
sin(θ) =
opposite
hypotenuse
cos(θ) =
adjacent
hypotenuse
tan(θ) =
opposite
adjacent
so we can easily find the ratios for these three angles:
sin
π
4
=
Finding
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Trigonometric ratios for certain angles
√
2
π
4
π
6
2
1
√
3
π
3
1
1
We know that, for right angle triangles,
sin(θ) =
opposite
hypotenuse
cos(θ) =
adjacent
hypotenuse
tan(θ) =
opposite
adjacent
so we can easily find the ratios for these three angles:
sin
π
4
=
Finding
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Workshop
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Trigonometric ratios for certain angles
√
2
π
4
π
6
2
1
√
3
π
3
1
1
We know that, for right angle triangles,
sin(θ) =
opposite
hypotenuse
cos(θ) =
adjacent
hypotenuse
tan(θ) =
opposite
adjacent
so we can easily find the ratios for these three angles:
sin
π
4
=
1
√
2
Finding
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Trigonometric ratios for certain angles
√
2
π
4
π
6
2
1
√
3
π
3
1
1
We know that, for right angle triangles,
sin(θ) =
opposite
hypotenuse
cos(θ) =
adjacent
hypotenuse
tan(θ) =
opposite
adjacent
so we can easily find the ratios for these three angles:
sin
π
4
=
1
√
2
cos
π
4
=
√1
2
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Trigonometric ratios for certain angles
√
2
π
4
π
6
2
1
√
3
π
3
1
1
We know that, for right angle triangles,
sin(θ) =
opposite
hypotenuse
cos(θ) =
adjacent
hypotenuse
tan(θ) =
opposite
adjacent
so we can easily find the ratios for these three angles:
sin
π
4
=
1
√
2
cos
π
4
=
√1
2
Finding
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)
tan
Contents
π
4
Prev
=
1
1
=1
Next
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Trigonometric ratios for certain angles
√
2
π
4
π
6
2
1
√
3
π
3
1
1
We know that, for right angle triangles,
sin(θ) =
opposite
hypotenuse
cos(θ) =
adjacent
hypotenuse
tan(θ) =
opposite
adjacent
so we can easily find the ratios for these three angles:
sin
π
4
sin
π
3
=
=
1
√
2
√
3
2
cos
π
4
cos
π
3
=
=
√1
2
1
2
Finding
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tan
π
4
tan
π
Contents
3
Prev
=
1
1
=1
√
=
3
1
Next
=
√
3
6 / 11
Trigonometric ratios for certain angles
√
2
π
4
π
6
2
1
√
3
π
3
1
1
We know that, for right angle triangles,
sin(θ) =
opposite
hypotenuse
cos(θ) =
adjacent
hypotenuse
tan(θ) =
opposite
adjacent
so we can easily find the ratios for these three angles:
sin
π
4
sin
π
sin
3
π
6
1
√
2
=
=
=
√
3
2
1
2
cos
π
4
cos
π
cos
3
π
6
=
=
√1
2
1
2
π
4
tan
π
tan
π
6
√
=
3
2
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tan
Contents
3
Prev
=
1
1
=1
√
=
3
1
=
√1
3
Next
=
√
3
6 / 11
Trigonometric ratios for angles >
So, what about angles such as
3π
4
π
2
which can’t be found in right angle triangles?
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Trigonometric ratios for angles >
So, what about angles such as
3π
4
π
2
which can’t be found in right angle triangles?
To extend trig ratios, we use a circle of radius 1 centred on the origin in the x − y plane
(the unit circle).
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Trigonometric ratios for angles > π2 : the unit circle
1
−1
1
−1
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Trigonometric ratios for angles > π2 : the unit circle
1
−1
1
−1
Start with a point P on the circle.
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Trigonometric ratios for angles > π2 : the unit circle
1
P (x , y )
−1
1
−1
Start with a point P on the circle.
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Trigonometric ratios for angles > π2 : the unit circle
1
P (x , y )
−1
1
−1
Start with a point P on the circle.
P has co-ordinates (x , y ).
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Trigonometric ratios for angles > π2 : the unit circle
1
P (x , y )
−1
1
−1
Start with a point P on the circle.
P has co-ordinates (x , y ).
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Trigonometric ratios for angles > π2 : the unit circle
1
P (x , y )
−1
1
−1
Start with a point P on the circle.
P has co-ordinates (x , y ).
This defines a right angle triangle
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Trigonometric ratios for angles > π2 : the unit circle
1
P (x , y )
−1
1
−1
Start with a point P on the circle.
P has co-ordinates (x , y ).
This defines a right angle triangle
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Trigonometric ratios for angles > π2 : the unit circle
1
P (x , y )
−1
1
−1
Start with a point P on the circle.
P has co-ordinates (x , y ).
This defines a right angle triangle with the side lengths shown.
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Trigonometric ratios for angles > π2 : the unit circle
1
P (x , y )
1
y
x
−1
1
−1
Start with a point P on the circle.
P has co-ordinates (x , y ).
This defines a right angle triangle with the side lengths shown.
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Trigonometric ratios for angles > π2 : the unit circle
1
P (x , y )
1
y
x
−1
1
−1
Start with a point P on the circle.
P has co-ordinates (x , y ).
This defines a right angle triangle with the side lengths shown.
Call the angle at the origin θ.
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Trigonometric ratios for angles > π2 : the unit circle
1
P (x , y )
1
y
θ
x
−1
1
−1
Start with a point P on the circle.
P has co-ordinates (x , y ).
This defines a right angle triangle with the side lengths shown.
Call the angle at the origin θ.
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Trigonometric ratios for angles > π2 : the unit circle
1
P (x , y )
1
y
θ
x
−1
1
−1
Since the hypotenuse is 1, we can link the angle (θ) to the coordinates of P:
sin(θ) = y
and
cos(θ) = x
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Trigonometric ratios for angles > π2 : the unit circle
1
P (x , y )
1
y
θ
x
−1
1
−1
Since the hypotenuse is 1, we can link the angle (θ) to the coordinates of P:
sin(θ) = y
and
cos(θ) = x
These relationships hold for any angle θ.
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Trigonometric ratios for angles > π2 : the unit circle
1
P
3π
4
−1
1
−1
Since the hypotenuse is 1, we can link the angle (θ) to the coordinates of P:
sin(θ) = y
and
cos(θ) = x
These relationships hold for any angle θ.
Example: Here is the situation for θ =
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)
3π
.
4
Contents
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Trigonometric ratios for angles > π2 : the unit circle
1
P
3π
4
−1
1
−1
Notice how this angle relates to the “first quadrant” angle
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π
.
4
Prev
Next
8 / 11
Trigonometric ratios for angles > π2 : the unit circle
1
P
3π
4
−1
1
−1
Notice how this angle relates to the “first quadrant” angle
π
.
4
P has the same y −coordinate so
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Trigonometric ratios for angles > π2 : the unit circle
1
P
3π
4
−1
1
−1
Notice how this angle relates to the “first quadrant” angle
P has the same y −coordinate so sin
3π
4
= sin
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π
4
π
.
4
Contents
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Trigonometric ratios for angles > π2 : the unit circle
1
P
3π
4
−1
1
−1
Notice how this angle relates to the “first quadrant” angle
P has the same y −coordinate so sin
3π
4
= sin
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π
4
Contents
=
π
.
4
√1
2
Prev
Next
8 / 11
Trigonometric ratios for angles > π2 : the unit circle
1
P
3π
4
−1
1
−1
Notice how this angle relates to the “first quadrant” angle
P has the same y −coordinate so sin
3π
4
= sin
π
4
=
π
.
4
√1
2
P’s x −coordinate is negative so
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Trigonometric ratios for angles > π2 : the unit circle
1
P
3π
4
−1
1
−1
Notice how this angle relates to the “first quadrant” angle
P has the same y −coordinate so sin
P’s x −coordinate is negative so cos
3π
4
= sin
π
4
=
= − cos
π
4
= − √12
Contents
Prev
3π
4
π
.
4
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)
√1
2
Next
8 / 11
Trigonometric ratios for angles > π2 : the unit circle
1
−1
1
−1
Example: Find sin
4π
3
and cos
4π
3
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Trigonometric ratios for angles > π2 : the unit circle
1
−1
1
−1
4π
3
and cos
4π
3
1. Sketch a unit circle and roughly locate P for θ =
4π
.
3
Example: Find sin
Strategy:
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Trigonometric ratios for angles > π2 : the unit circle
1
4π
3
−1
P
1
−1
4π
3
and cos
4π
3
1. Sketch a unit circle and roughly locate P for θ =
4π
.
3
Example: Find sin
Strategy:
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Trigonometric ratios for angles > π2 : the unit circle
1
4π
3
−1
P
1
−1
Example: Find sin
4π
3
and cos
4π
3
Strategy:
4π
.
3
angle π3
1. Sketch a unit circle and roughly locate P for θ =
2. Relate it to the appropriate “first quadrant”
Finding
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Workshop
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Semester
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angles
)
.
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Trigonometric ratios for angles > π2 : the unit circle
1
4π
3
−1
P
1
−1
Example: Find sin
4π
3
and cos
4π
3
Strategy:
4π
.
3
angle π3
1. Sketch a unit circle and roughly locate P for θ =
2. Relate it to the appropriate “first quadrant”
Finding
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Workshop
ratios
Semester
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2, 2016
angles
)
.
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Trigonometric ratios for angles > π2 : the unit circle
1
4π
3
−1
P
1
−1
Example: Find sin
4π
3
and cos
4π
3
Strategy:
4π
.
3
angle π3
1. Sketch a unit circle and roughly locate P for θ =
2. Relate it to the appropriate “first quadrant”
3. Work out (or recall) the trig ratios for
π
3
sin
π
3
.
√
=
Finding
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Workshop
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Semester
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2, 2016
angles
)
3
,
2
cos
Contents
π
3
Prev
=
1
2
.
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Trigonometric ratios for angles > π2 : the unit circle
1
4π
3
−1
P
1
−1
Example: Find sin
4π
3
and cos
4π
3
Strategy:
4π
.
3
angle π3
1. Sketch a unit circle and roughly locate P for θ =
2. Relate it to the appropriate “first quadrant”
3. Work out (or recall) the trig ratios for
4. Change to negative if required.
sin
π
3
4π
3
sin
π
3
.
√
=
3
,
2
cos
π
3
=
1
2
.
=
Finding
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trigonometric
Workshop
ratios
Semester
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2, 2016
angles
)
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Trigonometric ratios for angles > π2 : the unit circle
1
4π
3
−1
P
1
−1
Example: Find sin
4π
3
and cos
4π
3
Strategy:
4π
.
3
angle π3
1. Sketch a unit circle and roughly locate P for θ =
2. Relate it to the appropriate “first quadrant”
3. Work out (or recall) the trig ratios for
4. Change to negative if required.
sin
π
3
4π
3
sin
π
3
.
√
=
3
,
2
cos
π
3
=
1
2
.
√
=− 23 ,
Finding
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trigonometric
Workshop
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Semester
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2, 2016
angles
)
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Trigonometric ratios for angles > π2 : the unit circle
1
4π
3
−1
P
1
−1
Example: Find sin
4π
3
and cos
4π
3
Strategy:
4π
.
3
angle π3
1. Sketch a unit circle and roughly locate P for θ =
2. Relate it to the appropriate “first quadrant”
3. Work out (or recall) the trig ratios for
4. Change to negative if required.
sin
π
3
4π
3
π
.
√
=
3
,
2
√
=− 23 , cos
4π
3
sin
3
Finding
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trigonometric
Workshop
ratios
Semester
for certain
2, 2016
angles
)
cos
Contents
π
3
=
1
2
.
=
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8 / 11
Trigonometric ratios for angles > π2 : the unit circle
1
4π
3
−1
P
1
−1
Example: Find sin
4π
3
and cos
4π
3
Strategy:
4π
.
3
angle π3
1. Sketch a unit circle and roughly locate P for θ =
2. Relate it to the appropriate “first quadrant”
3. Work out (or recall) the trig ratios for
4. Change to negative if required.
sin
π
3
4π
3
sin
π
3
.
√
=
√
=− 23 , cos
Finding
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trigonometric
Workshop
ratios
Semester
for certain
2, 2016
angles
)
3
,
2
4π
3
cos
Contents
π
3
=
1
2
.
=− 21 .
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Solving trigonometric equations
Solve for θ : cos(θ) = − 21 on the domain θ [0, 3π].
Finding
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Workshop
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Semester
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2, 2016
angles
)
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Solving trigonometric equations
Solve for θ : cos(θ) = − 21 on the domain θ [0, 3π].
1. Sketch a unit circle and find a point with x −coordinate − 12 (cosine ratio).
1
−1
1
−1
Finding
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Workshop
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Semester
for certain
2, 2016
angles
)
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Solving trigonometric equations
Solve for θ : cos(θ) = − 21 on the domain θ [0, 3π].
1. Sketch a unit circle and find a point with x −coordinate − 12 (cosine ratio).
1
−1
1
−1
Finding
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trigonometric
Workshop
ratios
Semester
for certain
2, 2016
angles
)
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Solving trigonometric equations
Solve for θ : cos(θ) = − 21 on the domain θ [0, 3π].
1. Sketch a unit circle and find a point with x −coordinate − 12 (cosine ratio).
2. Relate this point to the appropriate “first quadrant” angle.
1
−1
1
−1
Finding
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trigonometric
Workshop
ratios
Semester
for certain
2, 2016
angles
)
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9 / 11
Solving trigonometric equations
Solve for θ : cos(θ) = − 21 on the domain θ [0, 3π].
1. Sketch a unit circle and find a point with x −coordinate − 12 (cosine ratio).
2. Relate this point to the appropriate “first quadrant” angle.
1
−1
1
−1
Finding
((mα+hs)Smart
trigonometric
Workshop
ratios
Semester
for certain
2, 2016
angles
)
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9 / 11
Solving trigonometric equations
Solve for θ : cos(θ) = − 21 on the domain θ [0, 3π].
1. Sketch a unit circle and find a point with x −coordinate − 12 (cosine ratio).
2. Relate this point to the appropriate “first quadrant” angle.
3. We know that cos
π
3
=
1
2
1
−1
1
−1
Finding
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trigonometric
Workshop
ratios
Semester
for certain
2, 2016
angles
)
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9 / 11
Solving trigonometric equations
Solve for θ : cos(θ) = − 21 on the domain θ [0, 3π].
1. Sketch a unit circle and find a point with x −coordinate − 12 (cosine ratio).
2. Relate this point to the appropriate “first quadrant” angle.
3. We know that cos
π
3
=
1
2
1
π
3
−1
1
−1
Finding
((mα+hs)Smart
trigonometric
Workshop
ratios
Semester
for certain
2, 2016
angles
)
Contents
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9 / 11
Solving trigonometric equations
Solve for θ : cos(θ) = − 21 on the domain θ [0, 3π].
1. Sketch a unit circle and find a point with x −coordinate − 12 (cosine ratio).
2. Relate this point to the appropriate “first quadrant” angle.
3. We know that cos
π
3
=
1
2
so θ = π −
π
3
=
2π
3
is a solution.
1
π
3
−1
1
−1
Finding
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trigonometric
Workshop
ratios
Semester
for certain
2, 2016
angles
)
Contents
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9 / 11
Solving trigonometric equations
Solve for θ : cos(θ) = − 21 on the domain θ [0, 3π].
1. Sketch a unit circle and find a point with x −coordinate − 12 (cosine ratio).
2. Relate this point to the appropriate “first quadrant” angle.
3. We know that cos
π
3
=
1
2
so θ = π −
π
3
=
2π
3
is a solution.
1
2π
3 π
3
−1
1
−1
Finding
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trigonometric
Workshop
ratios
Semester
for certain
2, 2016
angles
)
Contents
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9 / 11
Solving trigonometric equations
Solve for θ : cos(θ) = − 21 on the domain θ [0, 3π].
1. Sketch a unit circle and find a point with x −coordinate − 12 (cosine ratio).
2. Relate this point to the appropriate “first quadrant” angle.
3. We know that cos
π
3
=
1
2
so θ = π −
π
3
=
2π
3
is a solution.
4. Clearly, there is another solution.
1
2π
3 π
3
−1
1
−1
Finding
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trigonometric
Workshop
ratios
Semester
for certain
2, 2016
angles
)
Contents
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9 / 11
Solving trigonometric equations
Solve for θ : cos(θ) = − 21 on the domain θ [0, 3π].
1. Sketch a unit circle and find a point with x −coordinate − 12 (cosine ratio).
2. Relate this point to the appropriate “first quadrant” angle.
3. We know that cos
π
3
=
1
2
so θ = π −
π
3
=
2π
3
is a solution.
4. Clearly, there is another solution.
1
2π
3 π
3
−1
1
−1
Finding
((mα+hs)Smart
trigonometric
Workshop
ratios
Semester
for certain
2, 2016
angles
)
Contents
Prev
Next
9 / 11
Solving trigonometric equations
Solve for θ : cos(θ) = − 21 on the domain θ [0, 3π].
1. Sketch a unit circle and find a point with x −coordinate − 12 (cosine ratio).
2. Relate this point to the appropriate “first quadrant” angle.
3. We know that cos
π
3
=
1
2
so θ = π −
π
3
4. Clearly, there is another solution. θ = π +
2π
is
3
π
4π
= 3.
3
=
a solution.
1
2π
3 π
3
−1
1
−1
Finding
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trigonometric
Workshop
ratios
Semester
for certain
2, 2016
angles
)
Contents
Prev
Next
9 / 11
Solving trigonometric equations
Solve for θ : cos(θ) = − 21 on the domain θ [0, 3π].
1. Sketch a unit circle and find a point with x −coordinate − 12 (cosine ratio).
2. Relate this point to the appropriate “first quadrant” angle.
3. We know that cos
π
3
=
1
2
so θ = π −
π
3
4. Clearly, there is another solution. θ = π +
2π
is
3
π
4π
= 3.
3
=
a solution.
1
2π
3 π
3
−1
4π
3
1
−1
Finding
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trigonometric
Workshop
ratios
Semester
for certain
2, 2016
angles
)
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9 / 11
Solving trigonometric equations
Solve for θ : cos(θ) = − 12 on the domain θ [0, 3π].
1
2π
3
−1
4π
3
1
−1
Finding
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trigonometric
Workshop
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Semester
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2, 2016
angles
)
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10 / 11
Solving trigonometric equations
Solve for θ : cos(θ) = − 12 on the domain θ [0, 3π].
1
2π
3
−1
4π
3
1
−1
5. Note that the domain is [0, 3π], which is more than a full circle.
Finding
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trigonometric
Workshop
ratios
Semester
for certain
2, 2016
angles
)
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10 / 11
Solving trigonometric equations
Solve for θ : cos(θ) = − 12 on the domain θ [0, 3π].
1
2π
3
−1
4π
3
1
−1
5. Note that the domain is [0, 3π], which is more than a full circle. This means that
there is another solution:
Finding
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trigonometric
Workshop
ratios
Semester
for certain
2, 2016
angles
)
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10 / 11
Solving trigonometric equations
Solve for θ : cos(θ) = − 12 on the domain θ [0, 3π].
1
2π
3
−1
4π
3
1
−1
5. Note that the domain is [0, 3π], which is more than a full circle. This means that
there is another solution:
Finding
((mα+hs)Smart
trigonometric
Workshop
ratios
Semester
for certain
2, 2016
angles
)
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10 / 11
Solving trigonometric equations
Solve for θ : cos(θ) = − 12 on the domain θ [0, 3π].
1
2π
3
−1
4π
3
1
−1
5. Note that the domain is [0, 3π], which is more than a full circle. This means that
there is another solution: 2π + 2π
= 8π
.
3
3
Finding
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Workshop
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Semester
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2, 2016
angles
)
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10 / 11
Solving trigonometric equations
Solve for θ : cos(θ) = − 12 on the domain θ [0, 3π].
1
2π
3
−1
4π
3
1
−1
5. Note that the domain is [0, 3π], which is more than a full circle. This means that
there is another solution: 2π + 2π
= 8π
.
3
3
6. The complete solution is θ =
2π 4π 8π
, 3, 3.
3
Finding
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Workshop
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Semester
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)
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Using STUDYSmarter Resources
This resource was developed for UWA students by the STUDYSmarter team for the
numeracy program. When using our resources, please retain them in their original form
with both the STUDYSmarter heading and the UWA crest.
Finding
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Workshop
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Semester
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2, 2016
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)
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