Trigonometry
Trigonometric Graphs
Trigonometric Graphs
Stephen Perencevich
Trigonometry
Trigonometric Graphs
Stephen Perencevich
Georg Cantor Institute for
Mathematical Studies
Silver Spring, MD
scp1usa@gmail.com
c
!2011
All rights reserved.
Trigonometry
Trigonometric Graphs
If θ is an angle in standard position and (x, y) is a point on the terminal ray of θ, and r =
then
p
x2 + y 2 ,
by definition:
sin t =
y
r
cos t =
x
r
y
We oberve that if r = 1, that is that if the point (x, y) has distance 1 to the origin, then sin θ = = y
1
x
and cos θ = = x, so that (x, y) = (cos θ, sin θ). The set of all points in the plane that have distance 1 to
1
the origin form a circle of radius 1, centered at the origin; this circle is called the unit circle.
The coordinates of the point of intersection
of the terminal ray of θ and the unit circle are
(cos θ, sin θ). This is sometimes taken as the definition of the sine and cosine function.
Instead of applying sine and cosine to single angles as we would when solving a triangle, we wish to
view sine and cosine as functions with angles as inputs and construct their graphs. To construct the sine
function we begin with the terminal ray for 0 we rotate counterclockwise through 2π radians and track the
y-coordinate of the point of intersection of the terminal ray and the unit circle.
Trigonometry
Trigonometric Graphs
On the diagrams below the graph of y(θ) = sin θ is shown to the right of the unit circle. Several values of
θ are selected and the points (θ, sin θ) are marked on the graph of the sine function. Note that as θ increases
from 0 to π/2, sin θ increases from 0 to 1. As θ increases from π/2 to π, sin θ decreases from 1 to 0. As θ
increases from π to 3π/2, sin θ becomes negative and decreases from 0 to −1. As θ increases from 3π/2 to
2π, sin θ increases from −1 to 0.
Trigonometry
Trigonometric Graphs
Here is the graph of y(θ) = sin θ with the θ axis labeled according to which quadrant contains the
terminal rays for a given interval of angles.
Angles of measures larger than 2π are constructed by continuing to rotate the terminal ray after the
first circuit, and angles of negative measure are constructed by rotating the terminal ray clockwise. The
terminal ray will trace out exactly the same points on the unit circle in these cases, so the values of the sine
function repeat themselves. This is just another way of saying that the sines of coterminal angles are equal.
Theorem The sine function is periodic with period 2π: sin(θ + 2π) = sin θ.
Because the sine function is periodic the graph of y(θ) = sin θ repeats the shape already drawn on the
interval [0, 2π].
Trigonometry
Trigonometric Graphs
A similar analysis leads to the shape of the cosine curve. The cosine of an angle is the x-coordinate
if the point of intersection of the terminal ray of the angle and the unit circle. So cos 0 = 1, cos π/2 = 0,
cos π = −1, cos 3π/2 = 0, and cos 2π = 1. The sine and cosine curves are shown below.
We turn now to the task of sketching some vatiations of the functions y(θ) = sin θ and x(θ) = cos θ
obtained principally by replacing θ with an expression in t, which we regard as measuring time in seconds.
It is helpful to visualize the terminal ray rotating about the origin. With this in mind, if we simply set θ = t
we obtain x(t) = cos t and y(t) = sin t which give the x- and y-coordinates of the intersection point of the
terminal ray and the unit circle, after t seconds, where the terminal ray makes a circuit every 2π ≈ 6.28
seconds. We will proceed with examples.
Example Sketch the graph of y(t) = sin 2t.
This modification corresponds to setting θ = 2t. When t = π, θ = 2π, so terminal ray makes a circuit
in half the time. The period of this modified function is π . To complete the graph we must find the
t-coordinates of the maximum and minimum points and of the t-intercepts.
P2 is the half way point of the period, and so has
1
π
t-coordinate π = .
2
2
1 π
π
The t-coordinate of P1 is · = .
2 2
4
P3 is the midpoint between P2 and (π.0), so the tcoordinate can be found by computint the average:
π
3π
+π
3π 1
3π
2
2
=
=
· =
2 2
4
2
2
Trigonometry
Trigonometric Graphs
When skecthed by itself, the graph of y(t) = sin 2t
looks identical to the graph of y(t) = sin t, but the difference is clear when they are sketched on the same set of
axes.
Example Sketch the graph of y(t) = sin 2πt.
Again we have changed the period of the curve by setting θ = 2πt. The rotating terminal ray will
complete a circuit when θ = 2π. Solving the equation 2πt = 2π yields t = 1, so a circuit is completed
every second, and y(t) = sin 2πt has period 1. To complete the graph we must find the t-coordinates of the
maximum and minimum points and of the t-intercepts.
The t-coordinates of the maximum and minimum
points together with the t coordinate of the t-intercept
divide the period interval into four equal-length intervals, so it is clear that P1 = 1/4, P2 = 1/2, and
P3 = 3/4
In general if θ is replaced by bt, for some b to form the functions y(t) = sin bt or x(t) = cos bt, θ = 2π
when bt = 2π ⇒ b = 2π/b. We have thus obtained the following rule:
2π
b
2π
The period of x(t) = cos bt is
b
The period of y(t) = sin bt is
Exercise Find the period of the following functions.
1. y(t) = sin 3t
2. x(t) = cos πt
3. y(t) = sin 12 t
4. x(t) = cos 4t
5. x(t) = cos 3πt
6. y(t) = sin 6t
Trigonometry
Trigonometric Graphs
If (x, y) is a point on the terminal ray of angle θ and r =
p
x2 + y 2 = 2, then
sin θ = y/2 ⇒ y = 2 sin θ
cos θ = x/2 ⇒ x = 2 cos θ
The set of all points with r = 2 (The distance to the origin is 2.) forms a circle of radius 2 centered at
the origin. The intersection of the terminal ray of the angle θ with this circle has coordinates (2 cos θ, 2 sin θ).
Setting θ = t, we again imagine the terminal ray rotating about the origin, making a circuit every
2π seconds, so that x(t) = 2 cos t and y(t) = 2 sin t give the coordinates of the point of intersection of the
terminal ray with the circle of radius 2 after t seconds. The values of these functions are exactly double
those of cos t and sin t, so the graphs are stretched vertically.
There is nothing special about 2 in the previous example. The graphs of the functions x(t) = A cos t and
y(t) = A sin t reach a maximum height of A above the t-axis. A is called the amplitude of these functions.
Trigonometry
Trigonometric Graphs
We now consider one last modification to the function y(θ) = sin θ. Again we think of t as time measured
in seconds, and let θ = t − π/4, so the function we will analyze is y(t) = sin(t − π/4). Once again we consider
our terminal ray rotating around the origin. Since θ = t − π/4, θ = −π/4 when t = 0, so instead of beginning
its rotation from the positive x-axis (angle measure zero), our terminal ray begins its rotation from the
fourth quadrant at the angle −π/4.
We will sketch one period of the function y(t) = sin(t − π/4). Our angle θ is given by θ = t − π/4, and
we will sketch the period of the function betweem θ = 0 and θ = 2π. To this end we will solve tow equations
to find the tiimes that produce these angles.
Begin: θ = 0
End: θ = 2π
t − π/4 = 0
t − π/4 = 2π
t = π/4
t = 2π + π/4
t = 9π/4
To finish the sketch we need to find the locations of P1 , P2 , and P3 as indicated on the diagram.
P2 is the midpoint of the beginning and
edning point of the period interval, so we
can find its location by averaging:
P2 :
π/4 + 9π/4
=
10π/4
=
2
2
We find P1 and P3 similarly.
P1 :
P3 :
π/4 + 5π/4
2
3π
=
4
2
14π/4
7π
=
=
4
2
=
2
5π/4 + 9π/4
6π/4
5π
4
Trigonometry
Trigonometric Graphs
The graph of y(t) = sin(t − π/4) is obtained by translating the graph of
y(t) = sin t to the right by π/4.
A horizintal shift of this type is called a
phase shift and the angle π/4 is designated by ϕ (the Greek letter phi) and is
called the phase angle.
The most general form of the sine and cosine functions is as follows:
y(t) = A sin b(t − ϕ)
x(t) = A cos b(t −ϕ)
Amplitude= A
2π
b
Phase Angle= ϕ
Period=
One period of these graphs begins at the phase angle ϕ. The points P1 , P2 , and P3 divide the period
interval into four equal-length intervals.
Trigonometry
Trigonometric Graphs
Example Sketch one period of theh graph of x(t) = cos 3(t + π/3).
For this function we have set θ = 3(t + π/3). We can find the beginning and ending points of one period
of this function by solving the equations θ = 0 and θ = 2π.
Begin: θ = 0
End: θ = 2π
3(t + π/3) = 0
3(t + π/3) = 2π
t + π/3 = 0
t + π/3 = 2π/3
t = −π/3
t = 2π/3 − π/3
t = π/3
The phase angle of this function is ϕ = −π/3. As a drawing aid a vertical dashing line has been added
to the diagram at −π/3 to simulate the y-axis which will be added to the diagram after the locations of P1 ,
P2 , and P3 are determined.
P2 is the midpoint of the beginning and
edning point of the period interval, so we
can find its location by averaging:
P2 :
−π/3 + π/3
2
=0
We find P1 and P3 similarly.
P1 :
P3 :
−π/3 + 0
2
0 + π/3
2
=−
=
π
6
π
6
The y-axis passes through the point P1 .
Trigonometry
Trigonometric Graphs
Example Sketch one period of theh graph of y(t) = 3 sin(2t + π/3).
For this function we have set θ = 2t + π/3. We can find the beginning and ending points of one period
of this function by solving the equations θ = 0 and θ = 2π.
Begin: θ = 0
End: θ = 2π
2t + π/3 = 0
2t + π/3 = 2π
2t = −π/3
2t = 5π/3
t = −π/6
t = 5π/6
The phase angle of this function is ϕ = −π/6. As a drawing aid a vertical dashing line has been added
to the diagram at −π/6 to simulate the y-axis which will be added to the diagram after the locations of P1 ,
P2 , and P3 are determined.
P2 is the midpoint of the beginning and
edning point of the period interval, so we
can find its location by averaging:
P2 :
−π/6 + 5π/6
P1 :
−π/6 + π/3
4π/6
π
=
3
2
2
We find P1 and P3 similarly.
P3 :
2
π/3 + 5π/6
2
=
=
=
π/6
2
=
π
12
7π/6
7π
=
2
12
The y-axis passes between −π/6 and P1 .
Note that for this example the function is not quite in standard form. As it is written, θ = 2t + π/3.
This can be put in standard form by factoring 2 from each term: θ = 2t + π/3 = 2(t + π/6) = 2(t − (−π/6)),
so that y(t) = 3 sin 2(t − (−π/6)). We can now read directly from the equation:
Amplitude = A = 3
2π
2π
=
=π
b
2
Phase Angle = ϕ = −π/6
Period =
Note that regardless of the form of a sine or cosine function, the phase angle is always the solution of
the equation θ = 0. We can also find the period of this function by computing the length of the interval
on which the period of this function is graphed: 5π/6 − (−π/6) = π. The periods of all sine and cosine
functions can be computed similarly.
The tangent curve
If (x, y) are the coordinates of a point on the terminal ray of the angel θ, then, by definition, tan θ = y/x.
Taking (x, y) to be a point on the unit circle,
we have x = cos θ and y = sin θ, so the following
identity holds:
tan θ =
sin θ
cos θ
Since cos π/2 = 0 and cos 3π/2 = 0, the tangent function is not defined at these values. The tangent
function is also undefined at −π/2, since this angle is coterminal with 3π/2.
If θ is near π/2, then sin θ ≈ 1 and cos θ ≈ 0,
sin θ
so tan θ =
is very large. The same is true if
cos θ
θ is near −π/2, so the graph of the tangemt function has vertical asymptotes at θ = −π/2 and
θ = π/2.
f (θ) = tan θ is periodic with period π.
Just as with the sine and cosine functions, we are interested in sketching variations on the function
f (θ) = tan θ obtained by replacing θ with an expression in t.
Example Sketch the graph of f (t) = tan 3t.
For this example we have set θ = 3t. The tangent function has asymptotes when θ = −π/2 and when
θ = π/2; we solve two equations to find the locations of the asymptotes of the modified tangent function.
Begin: θ = −π/2
End: θ = π/2
3t = −π/2
3t = π/2
t = −π/6
t = π/6
Example Sketch the graph of f (t) = tan 2(t − π/6).
For this example we have set θ = 3t. The tangent function has asymptotes when θ = −π/2 and when
θ = π/2; we solve two equations to find the locations of the asymptotes of the modified tangent function.
Begin: θ = −π/2
End: θ = π/2
2(t − π/6) = −π/2
2(t − π/6) = π/2
t − π/6 = −π/4
t − π/6 = π/4
t = −π/4 + π/6
t = π/4 + π/6
t = −π/12
t = 5π/12
The t-intercept of the curve is the
midway between the asymptotes so we can
find it by averaging:
−π/12 + 5π/12
2
=
4π/12
2
=
π
6