Download The Unit Circle

Trigonometry
So far you work has been mainly restricted to using trigonometry for right-angled triangles.
Both acute angles i.e. < 90°
You have found the sine, cosine or tangent of acute angles in order to find unknown sides. You have
also used known sides to find the acute angles.
Trigonometric Graphs
From drawing the graphs of y = sin θ, y = cos θ and y = tan θ, you have discovered that sine, cosine
and tangent of larger angles also exists:
y
1
0.8
0.6
y = sinx
0.4
0.2
-360
-270
-180
-90
90
-0.2
-0.4
-0.6
-0.8
-1
180
270
360
x
y
1
0.8
0.6
0.4
y = cosx
0.2
-360
-270
-180
-90
90
180
270
x
360
-0.2
-0.4
-0.6
-0.8
-1
y
10
y = tanx
5
-360
-270
-180
-90
90
-5
-10
180
270
360
x
The sine and cosine graphs repeat their pattern every 360 degrees so sin (70°) = sin (70° + 360°), for
example. The tangent graph repeats its pattern every 180 degrees so tan (70°) = tan (70° + 180°).
What other things do you notice?
sin (-x) = - sin (x)
-1 ≤ sin θ ≤ 1
cos (-x) = cos (x)
-1 ≤ cos θ ≤ 1
tan (-x) = - tan (x)
The Unit Circle
1
2nd
Quadrant
-1
y
Angles are measured from the positive
x axis in an anticlockwise direction.
1st
Quadrant
3rd
Quadrant
4th
Quadrant
1
x
In 3A we will only concern
ourselves with the first two
quadrants:
240°
35°
0° ≤ θ ≤ 180°
-1
Finding the sine of an angle on the unit circle
The radius of a unit circle is 1
A
A is the coordinate point (x, y)
y
𝑦
θ
sin θ = 1
1
θ
y
i.e. sin θ = y
so sin θ is the y coordinate of the point A.
When the unit circle is drawn onto an accurate scale, the y coordinate can be estimated.
Sadler 3A Exercise 1A
It also works for angles between 90° and 180° (i.e. the second quadrant). Sin 135° is the y
coordinate of the point A below:
A (x, y)
y
1
45°
135°
sin 45° = y =
1
√2
1
45°
135°
on calculator sin 45° = 0.707 (also, on
calculator, sin 135° = 0.707).
The general rule is
sin (180° - θ) = sin θ
when
0° ≤ θ ≤ 90°
Finding the cosine of an angle on the unit circle
𝑥
A (x, y)
1
1
x
A(x, y)
Cos 70° = 1
Cos 70° = x
70°
x
So cos 70° is the x
coordinate of the point A
Cos 160° = the x coordinate
of the point A – WHICH IS A
NEGATIVE VALUE
1
160°
If 0° ≤ θ ≤ 90° then cos θ > 0
If 90° ≤ θ ≤ 180° then cos θ < 0
The following general rule applies:
Cos ( 180° - θ) = - cos θ
when 0° ≤ θ < 90°
Alternate Formula for the Area of a Triangle
We already know that:
h
h
b
h
b
1
2
Area =
×
base
×
b
perpendicular height
However, sometimes the perpendicular height is not given. Instead the lengths of the sides of the
triangle and some angles are given.
E.g.
B
B
a
A
C
b
Acute angled triangle
a
C
A
C
b
Obtuse angled triangle
If the known angle is between the two known sides then:
Area =
𝑎 ×𝑏 × sin 𝐶
2
See Sadler page 27 for the proof of this area formula and then
see page 28 for examples and complete Exercise 1B
Proof:
Finding θ, given sin θ, when 0° ≤ θ ≤ 180°
E.g. sin θ =
1
2
On Unit Circle:
1
Find 2 on the y-axis and draw a horizontal line to the circle.
0.5
Where it crosses the circle, join these points to the origin
(0, 0) and estimate the size of the angles.
Two possible angles are
30° and 150°
Remember that the unit circle is symmetrical so
Also remember that
150°
30°
180° - 30° = 150°
sin θ = sin (180° - θ )
30°
On the graph of y = sin θ:
y
1
0.5
30°
90
150°
180
x
There are e-activities for this area formula available for your
classpad. Ensure that you obtain a copy from your teacher.
Alternatively, try to write one yourself.
The Sine Rule for non-right triangles
It is not necessary to memorise this proof. However, it is worthwhile understanding it.
Examples:
Find the value of x in the following:
a)
70°
16.4 m
b)
120°
7.1 cm
60°
θ
x
𝑥
sin 70°
x =
=
10.3 cm
16.4
sin 60°
sin 𝜃
7.1
16 × sin 70°
sin 60°
=
sin 𝜃 =
sin 120
10.3
7.1 × sin 120°
10.3
𝜃 = sin−1 (
x = 17.79 m (2 d.p.)
7.1 × sin 120°
)
10.3
θ = 36.7° (1 d.p.)
Using solve facility on classpad
solve (
sin 𝜃
7.1
=
gives
and
sin 120
10.3
As angles in a triangle
add up to 180°
, θ ) ǀ 0 ≤ θ ≤ 180
θ = 36.65312298
θ = 143.346877
Why two answers?
There are always two angles between zero and 180 degrees that have the
same sine value. (Look back at the unit circles and the y coordinates)
However, we can discard 143.35° for this question.
120°
WHY?
We already have 120° so 143° is too big as angles
In a triangle add up to 180°.
Θ
See Sadler 3A Example 5 on page 33 and
then do Exercise 1C questions 1 to 9
The Cosine Rule
It is not necessary to memorise this proof. However, it is worthwhile understanding it.
c2 = a2 + b2 - 2ab cos C
→
cos 𝐶 =
𝑎2 + 𝑏2 − 𝑐 2
2𝑎𝑏
Examples:
θ
a)
14.6 cm
b)
x
5.2 m
6.9 m
50°
16.8 cm
𝑥 2 = 14.62 + 16.82 − 2 × 14.6 × 16.8 × cos 50°
8.3 m
cos 𝜃 =
5.22 +6.92 −8.32
2×5.2×6.9
𝑥 = √180.074
cos 𝜃 = 0.08
𝑥 = 13.42 𝑐𝑚 (2 𝑑. 𝑝. )
𝜃 = 85.4°
See Sadler Example 8 on page 36 and then do the rest of
Exercise 1C
Sadler 3A Misc Exercise 1