Download Background Notes on Trigonometry 1 Triangles and Circles

Background Notes on Trigonometry
Many real phenomena show a cyclic or repetitive type of behaviour. For example, business
cycles, sun spots, earthquakes, the motion of the planets, heartbeats and sound-waves, just
to name a few. In this topic we will study an important class of functions that are useful in
describing cyclic phenomena. They are called the trigonometric functions.
(“Trigon” comes from Greek roots meaning “three sides” and is another name for a triangle.
“-Metry” is a suffix meaning “measurement” and so the full term roughly translates to “triangle
measurement”.)
1
Triangles and Circles
Similar Triangles
If you have any triangle, and another one that looks identical except that it is either larger or
smaller (i.e. looks “blown-up” or “shrunk-down”), then those two triangles are similar . The
mathematical definition for similar triangles is that they both have corresponding angles that
are equal, while the lengths of the corresponding sides are in proportion.
F
C
E
B
A
D
|AC|
|AB|
|BC|
=
=
|DF |
|DE|
|EF |
and
d = Fd
CAB
DE ,
d = DEF
d ,
ABC
d = EF
dD
BCA
Comments
• We can use |AB| to denote the distance from point A to point B, but I’ll usually just
use a letter for the required length.
d to denote the measure of the angle formed by the line
• We can use 6 ABC or ABC
segments AB and BC, but again, I’ll usually use a (Greek) letter.
Example: Consider the similar triangles below.
10
a
b
7
3
6
Since the corresponding sides are in proportion,
a
b
3
=
=
7
10
6
So, a = 3.5 and b = 5.
1
(=
1
)
2
The Pythagorean Theorem
A right-angled triangle is one that contains a right angle. Note that the side of the triangle
opposite the right angle is called the hypotenuse.
Pythagoras’ Theorem: If a and b are the lengths of the sides of a right-angled triangle,
and c is the length of the hypotenuse, then
c
c2 = a2 + b2
That is,
b
a
“The square of the hypotenuse is equal to the sum of the squares of the sides”.
Example: Consider the following right-angled triangles.
5
(b)
(a)
4
12
x
1
(a) Pythagoras gives
x2 = 122 + 52
and taking the square root solves for the unknown x,
√
√
x = 144 + 25 = 169 = 13 .
(b) Pythagoras gives
42 = y 2 + 12 .
Rearranging and taking the square root solves for the unknown y,
y =
√
16 − 1 =
2
√
15 .
y
The Distance Between Two Points
Pythagoras’ Theorem leads to the following formula for finding the distance, d, between two
points P (x1 , y1 ) and Q(x2 , y2 ):
p
d =
(x2 − x1 )2 + (y2 − y1 )2 .
y
Q(x2 , y2 )
b
y2
d
P (x1 , y1 )
y1
b
y2 − y1
x2 − x 1
x
x1
x2
y
A quick look at Circles
r
The equation for a circle centered at
the origin with radius r is given by
b
x
x2 + y 2 = r 2 .
Comments:
• If we let r = 1, then we obtain the unit circle.
• When we are talking about circles, quite often we use “pi” denoted by π. It represents
the ratio of the circumference, C (the distance around the circle), to its diameter, d. (A
discovery made by the ancient Greeks!) Therefore we get
C = π d or
C = 2π r ,
since the radius is half the diameter.
Pi is an irrational number approximated by 3.141592, which is correct to six decimal
places.
• The area, A, of a circle of radius r is given by the formula A = π r 2 .
3
2
Angles and their Measure
Degree Measure of Angles
One common unit of measure of an angle is the degree.
If a circle is divided into 360 equal parts by radii, then the angle
between any two adjacent radii is one degree.
b
This means that there are 360◦ in one complete revolution.
Comments:
• A angle of measure 90◦ is called a right angle (as in Pythagoras’s Theorem).
• A angle of measure less than 90◦ is called an acute angle.
• One that measures greater than 90◦ is called an obtuse angle.
• The sum of all 3 angles in a triangle totals 180◦ .
Radian Measure of Angles
r
A very important unit of measure for angles is the radian.
One radian is the angle in a circle when the arc length, s,
is the same as the radius, r.
s=r
b
The relationship between degrees and radians is easy to figure out.
The circumference of the circle of radius 1 is 2π. So one complete revolution has a measure
of 2π radians, which we know is 360◦ . Therefore
360◦ = 2π radians,
In particular,
1 rad =
so, for example,
180◦
π
π
= 60◦
3
that is,
180◦ = π radians.
and 1◦ =
and
45◦ =
π
rad .
180
π
.
4
Arc Length and Area of a Sector
We can work out an arc length on a circle and the area
of a sector of the circle if we know the radius and the angle,
measured in radians, which the arc subtends.
If the angle measures θ radians, then
• the length of the arc is s = r θ , and
• the area of the sector is A =
1 2
r θ.
2
4
s
θ
b
r
3
The Trigonometric Functions
opposite
hy
p
ot
en
us
e
One of the easier ways to start understanding trigonometric functions is by picturing a righttriangle. Let θ be one of the acute angles and label the triangle as in the following diagram.
θ
adjacent
Now the three most common trigonometric functions, sine, cosine and tangent, can be defined
as the following ratios for any acute angle:
sin θ =
opposite
,
hypotenuse
cos θ =
adjacent
,
hypotenuse
tan θ =
opposite
sin θ
=
.
adjacent
cos θ
Others are:
csc θ =
hyp
1
=
,
opp
sin θ
sec θ =
hyp
1
=
,
adj
cos θ
cot θ =
adj
1
=
.
opp
tan θ
Comments:
• These ratios are the same for any right-triangle with a fixed angle θ, since when a triangle
has equal angles they are similar triangles.
• Your calculator will have built in trig. functions for angles both in degrees and radians.
For example, check using your calculator that
π
sin 30◦ = 0.5, cos = 0.707 and tan 60◦ = 1.732
4
(the last figures have been rounded to two decimal places).
Finding Sides
We can use these trigonometric ratios to find the length of any side in a right-triangle if we
know the length of one other side and one angle.
Example: Consider the triangle below with unknown lengths a and b.
Then, for example,
sin 30◦ =
so
a
12
12
a = 12 sin 30◦ = 6
30◦
b
The length b can be found in a similar way.
5
a
Finding Angles
Similarly we can find angles in a right-triangle if we know two of the sides. To do this we
need to use the inverse trigonometric functions, sin−1 , cos−1 and tan−1 to “undo” the effect
of taking the sin, cos and tan of an angle respectively. (Note: sin−1 x 6= sin1 x !!)
Example: Solving for α in the given triangle, we get
1
2
1
α = sin−1
= 30◦
2
sin α =
so
2
1
α
Two Special Triangles
You have probably noticed that you often need to round off calculator values. However, the
sine, cosine and tangent of certain angles can be found exactly, often in terms of surds, using
the following two triangles. The LH one is an isoceles triangle with two 45◦ angles and two
sides of length one. The RH one is half an equilateral triangle with 60◦ angles and all sides
of length two.
Isoceles Triangle
√
2
Equilateral Triangle
45◦
30◦
1
√
2
3
45◦
60◦
1
1
The above triangles can be used to complete the table below.
degrees
0◦
sin
0
cos
1
30◦
45◦
1
2
√
3
2
√1
2
√1
2
60◦
90◦
√
3
2
1
1
2
0
tan
0
√1
3
1
√
3
radians
0
π
6
π
4
π
3
∞
π
2
Note that the result sin 0 = 0 can be obtained by thinking of the degenerate case where a
“triangle” becomes a “line segment” so the “opposite” side has zero length. Similarly, cos 0 = 1
because the “adjacent” side will have the same length as the “hypotenuse” in this degenerate
case. We get tan 0 = 0 by noting that tan 0 = sin 0/ cos 0.
The ratios when the angle is 90◦ can be deduced by similar reasoning.
In fact, the ideas on the next page should help clarify this!
6
Angles larger than 90◦ .
We start with the unit circle centered at the origin and consider a point P (x, y) on the circle.
(The LHS figure shows some general point, and the RHS figure shows the point ( √12 , √12 )).
P (x, y)
y
b
√1
2
1
b
( √12 , √12 )
1
π
4
θ
√1
2
x
Since r = 1 we get
y
= y
1
and
cos θ =
π
1
= √
4
2
and
cos
sin θ =
x
= x.
1
For example,
sin
1
π
= √ ,
4
2
as we already know.
This means that the x-coordinate of the point P gives the value of cos and the y-coordinate
of P gives the value of sin. This is the way that the trig ratios are defined for angles larger
than 90◦ .
(Note that we could have chosen a circle with a different radius, but since the trig ratios are
the same as long as the angles are the same we might as well choose r = 1 for simplicity.)
For example, if θ =
sin
and
cos
3π
, then
4
3π
π
1
= sin
= √
4
4
2
(− √12 , √12 )
b
3π
π
1
= − cos
= −√
4
4
2
√1
2
3π
4
− √12
7
b
π
4
√1
2
( √12 , √12 )
Exercise
degrees
120◦
150◦
180◦
210◦
240◦
270◦
300◦
330◦
360◦
2π
3
5π
6
π
7π
6
4π
3
3π
2
5π
3
11π
6
2π
sin
cos
tan
radians
(You should check these results using your calculator.)
Comments
• Since x2 + y 2 = 1, we get the very important trigonometric identity
cos2 θ + sin2 θ = 1
for all angles θ.
• Another essential result we get from the previous diagram is
sin(θ + 2π) = sin θ
and
cos(θ + 2π) = cos θ
That is, the sine and cosine trigonometric functions are PERIODIC.
Comments
• The domain of both y = sin θ and y = cos θ is the reals, and the range is [−1, 1].
• The period , T , is 2π, since the sine and cosine curves repeat themselves every 2π radians.
8
4
The Graphs of Sine/Cosine/Tangent
Using the values in the previous two tables, and this periodic idea, we can obtain the following
key trigonometric graphs:
• y = sin θ.
y
y = sin θ
1
π
2
π
3π
2
2π
5π
2
θ
3π
2
2π
5π
2
θ
• y = cos θ.
y
y = cos θ
1
π
2
π
• y = tan θ.
y
y = tan θ
− π2
b
π
2
b
π
9
3π
2
θ
Comments
• From the graph we see both curves oscillate between 1 and -1. From this, we determine
the amplitude to be 1. (Subtract the lowest point from the highest point and divide
by 2.)
• Know what these graphs look like. In particular, it is easier to get the value for angles
greater that 90◦ from the symmetries of the graphs.
• More generally, if we consider sine and cosine curves of the form
y = a sin(b (θ − c)) + d and
y = a cos(b (θ − c)) + d
then the (absolute) value of a is the amplitude of the curve, d is the vertical shift and c
is the horizontal shift. (In the problems below, c = 0.)
The period of the curve is given by 2π/b.
We can use techniques similar to those used in plotting quadratics to graph sine/cosine curves.
Exercise: Sketch the graph of the following functions.
(a) y = sin(θ) + 1
(c) y = 2 sin(θ)
(b) y = sin(θ − π/2)
(d) y = sin(2θ)
Problem
A tsunami is a series of large waves caused by an earthquake. The wavelength of a tsunami can
be as long as several hundred kilometers. The figure below shows a tsunami in a co-ordinate
system. Write an equation for the tsunami assuming that its shape is that of a sinusoid.
y
Height (meters)
30
15
x
0
0
50
100
150
200
Wavelength (kilometers)
10
250
300