### Video Transcript

In this lesson, we will learn how
to graph trigonometric functions, such as sine and cosine, and deduce their
properties.

Weβll begin by looking at special
angles on the unit circle. This is a circle of radius one unit
whose center lies on the origin of the Cartesian plane. We might recall that the
π¦-coordinates of the various points that lie on this circle correspond to the sine
of the various angles. For instance, this point here tells
us that sin of 30 degrees is one-half, and this point tells us that sin of 45
degrees is root two over two.

Now, we can draw the given table to
relate the input, π₯, in degrees to the output for sin of π₯. Of course, since we can continue
moving around the circle infinitely and in either direction, we could extend this
table either way. This means that sine is a periodic
function. In particular, sin of π₯ has a
period of 360 degrees, or two π radians. A key feature of sin of π₯, which
is demonstrated in its graph, is that the function has a value of zero when π₯
equals zero degrees, and it increases to the maximum value one when π₯ equals 90
degrees.

By plotting the points from the
table above, we can approximate the graph of sin of π₯. As we saw, since the sine function
is periodic, we can extend it in either direction by repeating the graph over each
360-degree interval as shown here. We can now also see that the sine
function has roots; that is, it intersects the π₯-axis at every 180 degrees,
starting at zero. We can also see that thereβs
rotational symmetry about the origin; this means itβs an odd function. Letβs briefly summarize all of
this.

The graph of the sine function has
the following features. It has a π¦-intercept of zero and
has a maximum value of one and a minimum value of negative one. It has roots at every 180 degrees,
starting at zero. It is periodic with a period of 360
degrees or two π radians. Finally, itβs also an odd
function. Formally, that means that sin of
negative π₯ is equal to negative sin of π₯. Informally, it means it has
rotational symmetry about the origin.

Now, we can perform a similar
process to find coordinates on the graph of π¦ equals cos of π₯. This time, the π₯-coordinates of
the various points that lie on the unit circle correspond to the cosine of the
various angles. In fact, the inputβoutput table
looks like this. This gives us the graph of the
cosine function between zero and 360 degrees. Unlike the graph of sine, cosine
begins at the maximum value one when π₯ is zero degrees and decreases to the minimum
value negative one when π₯ is 180 degrees. Like sine, cosine is a periodic
function with a period of 360 degrees. And so, we can extend this graph
over a larger interval by making copies of the graph over the interval between zero
and 360 degrees.

Letβs summarize the key
features. The π¦-intercept of the cosine
function is one, which is its maximum, and it decreases to a minimum value of
negative one. It has roots every 180 degrees,
starting at 90. It has a period of 360 degrees or
two π radians. Finally, it is an even
function. In other words, cos of negative π₯
is equal to cos of π₯, and the graph is symmetric over the π¦-axis.

We will now consider our first
example in which we will look at how to use these properties to help us recognize
the graphs of these functions.

Which of the following is the graph
of π¦ equals cos of π₯?

Remember, one of the key features
of the cosine graph is that it has a π¦-intercept of one. This means we can immediately
disregard any graphs that donβt pass through the point zero, one. So, we can get rid of options (B),
(D), and (E) straightaway. Next, we know that itβs a periodic
function and that it repeats every 360 degrees. Option (C) appears to have a much
smaller period; in fact, it repeats every 120 degrees, so it canβt be this one.

This leaves only option (A). Letβs check by looking at the other
properties. Some of the roots of π¦ equals cos
of π₯ are 90 degrees, 270 degrees, and negative 90 degrees. We can see that option (A) passes
through all of these values on the π₯-axis. The curve has maxima at one and
minima at negative one. It is also an even function, so
itβs symmetric over the π¦-axis. Hence, we have confirmed that
option (A) is indeed the graph of π¦ equals cos of π₯.

In this example, we saw that it is
possible to identify the graph of cos of π₯ centered at the origin from its features
such as its π¦-intercept and its periodicity. The same principles apply to graphs
of cosine and sine when viewed at values of π₯ shifted away from the origin. In particular, we can use the
periodic nature of these functions to help us determine the location of key features
of the graph. Letβs demonstrate this in our next
example.

Consider the following figures. Part (1) Which function does the
plot in the graph, figure (a), represent? Is it (a) cosine or (b) sine? Part (2) Assign each region of the
plot in figure (a) to the corresponding quadrant of the unit circle in figure
(b).

To answer part (1), letβs recap on
the values of some of the coordinates of the graphs of the sine and cosine
functions. The coordinates of points on the
unit circle are given by cos π, sin π, where π is the counterclockwise angle of
the radius to that point measured from the positive π₯-axis. Specifically, in the given graph,
we can see that the value of the function is zero when the angle is two π radians;
thatβs a full revolution. The coordinate of the point on the
unit circle that corresponds to an angle of two π is one, zero, which means that
cos of two π is one and sin of two π is zero. The given graph indicates that this
function takes the value of zero at two π, so this agrees with the sine
function. This is option (B).

To answer part (2), we can use the
values of the π¦-coordinates on the unit circle. Letβs look at region A which lies
between three π over two and two π, where two π is a full revolution and three π
over two is three-quarters of this. That means that region A must lie
here, in the fourth quadrant. In a similar way, five π over two
must be a quarter turn further than a full turn. So, region B is here, in the first
quadrant. Region C is between five π over
two and three π, one and a half turns, so itβs the second quadrant, meaning that
region D must be in the third quadrant. So, we assign A to quadrant IV, B
to quadrant I, C to quadrant II, and D to quadrant III.

So far, we have only looked at the
graphs of π¦ equals sin of π₯ and π¦ equals cos of π₯. So, what happens if we multiply one
of these by some constant real number? For instance, take the function π¦
equals two cos of π₯. This means we input the value of π₯
into the function π¦ equals cos of π₯ then multiply that result by two. This means that all of the values
of π¦ would double. So, the maxima would now be two,
and the minima would now be negative two. So, multiplying the entire cosine
function by two gives us a vertical stretch by that scale factor. In fact, we will achieve a similar
result when multiplying the sine and cosine functions by a single constant
value. Letβs look in particular at what
happens if we multiply by a different constant.

Consider the graph below. Which function does the plot in the
graph represent? Is it option (A) π¦ equals cos of
π₯, (B) π¦ equals two cos of π₯, (C) π¦ equals negative sin π₯, (D) π¦ equals sin
π₯, or (E) π¦ equals negative cos π₯?

In this example, we have been given
a graph and need to decide which of the options represents it. Since all of the options include
sine or cos or constant multiples of these, we should begin by looking at the
features that these functions have. Letβs first recall the
π¦-intercepts of sine and cosine. The sine graph passes through the
π¦-axis at zero, whilst the cosine graph passes through at one. Comparing this to the graph given,
we see that the π¦-intercept is at zero, meaning that π¦ equals cos of π₯ cannot be
an option. In fact, we can see that this also
applies to the other options that are multiples of cos of π₯.

In particular, the graph of π¦
equals two cos π₯ is found by multiplying the values of the output of cos of π₯ by
two. This means the π¦-intercept would
be one times two, which is two. Similarly, the π¦-intercept of π¦
equals negative cos of π₯ is one times negative one, which is negative one. So, we can disregard options (B)
and (E) too. Letβs look at the remaining two
options.

Both of the graphs will pass
through the π¦-axis at zero, since negative one times zero is zero. We also know some of the key values
for the sine function. For the equation π¦ equals sin π₯,
when π₯ is equal to π by six, π¦ is equal to 0.5. And when π₯ is equal to π by two,
π¦ is equal to one. Our graph passes through the points
π by six, negative 0.5 and π by two, negative one, which means the output values
are being multiplied by negative one. To achieve this, we need to
multiply sin of π₯ by negative one. So, the graph has equation π¦
equals negative sin of π₯. Thatβs option (C).

Letβs now recap the key points from
this lesson. The graphs of the sine and cosine
functions are periodic, and each function has a period of 360 degrees or two π
radians. We saw that both have maxima and
minima at one and negative one, respectively. The π¦-intercept of the sine curve
is zero, whilst for the cosine curve itβs one. Finally, we learned that sine is an
odd function and cosine is even.