### Video Transcript

In this video, we will learn how to
simplify trigonometric expressions by applying trigonometric identities.

We begin by recalling that an
identity is an equation that is true no matter what values are chosen. We will use combinations of these
trigonometric identities, for example, the cofunction identities, shift identities,
and Pythagorean identities. Before looking at some specific
examples, letβs consider the properties of the unit circle.

We recall that the unit circle is a
circle of radius one as shown. It enables us to measure the sin,
cos, or tan of any angle π, where π is measured in the counterclockwise direction
from the positive π₯-axis. The π₯-coordinate of any point on
the unit circle is equal to cos π, and the π¦-coordinate is equal to sin π. The right triangle on our diagram
together with the Pythagorean theorem leads us to the first Pythagorean
identity. sin squared π plus cos squared π
is equal to one.

Recalling the reciprocal
trigonometric identities enables us to form two further Pythagorean identities. The three reciprocal functions are
the cosecant, secant, and cotangent such that csc π is equal to one over sin π,
sec π is equal to one over cos π, and cot π is equal to one over tan π. It is also worth noting that since
tan π is equal to sin π over cos π, then cot π is equal to cos π over sin
π. Dividing both sides of our first
Pythagorean identity by cos squared π, we have sin squared π over cos squared π
plus cos squared π over cos squared π is equal to one over cos squared π. Using the reciprocal identities,
this simplifies to tan squared π plus one is equal to sec squared π.

In the same way, we can divide both
sides of the first identity by sin squared π. This simplifies to one plus cot
squared π is equal to csc squared π. We now have a set of three
Pythagorean identities, which we will use together with the reciprocal identities to
solve a couple of examples.

Simplify sin π multiplied by csc
π minus cos squared π.

In this question, we are asked to
simplify a trigonometric expression. One way of doing this is using the
reciprocal and Pythagorean identities. In questions of this type, it is
not always clear what to do first. However, as a general rule, it is
worth replacing any reciprocal functions with the sine, cosine, or tangent
function.

We know that csc π is equal to one
over sin π. Substituting this into our
expression, we have sin π multiplied by one over sin π minus cos squared π. The sin π on the numerator and
denominator of our first term cancels, leaving us with one minus cos squared π. Next, we recall one of the
Pythagorean identities: sin squared π plus cos squared π is equal to one. Subtracting cos squared π from
both sides, this can be rewritten as sin squared π is equal to one minus cos
squared π. This means that our expression can
be rewritten as sin squared π. sin π multiplied by csc π minus
cos squared π written in its simplest form is sin squared π.

We will now consider a second
example of this type.

Simplify sin squared π plus cos
squared π divided by csc squared π minus cot squared π.

In order to answer this question,
we need to recall the Pythagorean identities. Firstly, we have sin squared π
plus cos squared π is equal to one. Dividing each term by sin squared
π and using our knowledge of the reciprocal trigonometric functions, we have one
plus cot squared π is equal to csc squared π. We notice that the left-hand side
of the first identity is identical to the numerator of our expression. Subtracting cot squared π from
both sides of the second identity, we have one is equal to csc squared π minus cot
squared π. The right-hand side of this is the
same as the denominator of our expression.

As sin squared π plus cos squared
π equals one and csc squared π minus cot squared π also equals one, our
expression simplifies to one divided by one. And this is equal to one. The expression sin squared π plus
cos squared π over csc squared π minus cot squared π is equal to one.

Before looking at one final
example, we will recall the cofunction and shift identities. Once again, we begin by considering
the unit circle. Since the angles in a triangle sum
to 180 degrees, the third angle in our right triangle is equal to 90 degrees minus
π. Letβs consider what happens if we
redraw this triangle such that the angle between the positive π₯-axis and the
hypotenuse is 90 degrees minus π. The coordinates of the point marked
on the unit circle will be cos of 90 degrees minus π, sin of 90 degrees minus
π. We notice that the distance in the
π₯-direction here is the same as the distance in the π¦-direction in our first
triangle. This means that the cos of 90
degrees minus π must be equal to sin π. Likewise, the sin of 90 degrees
minus π is equal to cos π.

Since sin π over cos π is equal
to tan π, the tan of 90 degrees minus π is equal to cos π over sin π. And using our knowledge of the
reciprocal functions, this is equal to cot π. It follows that the sec of 90
degrees minus π is equal to csc π. The csc of 90 degrees minus π is
equal to sec π. And the cot of 90 degrees minus π
is equal to tan π. These six identities are known as
the cofunction identities.

We can also use the unit circle to
find identities involving angles such as 180 degrees minus π, 180 degrees plus π,
and 360 degrees minus π. In our final example, we will use
these identities together with the Pythagorean identities to simplify an
expression.

Simplify one plus cot squared three
π over two minus π over one plus tan squared π over two minus π.

In order to answer this question,
we will need to use a variety of trigonometric identities. There are many ways to start
here. However, we will begin by trying to
rewrite the expression simply in terms of π. By firstly sketching the unit
circle, we recall that π radians is equal to 180 degrees. This means that π over two radians
is equal to 90 degrees. The denominator of our expression
can therefore be rewritten as one plus tan squared of 90 degrees minus π. One of our cofunction identities
states that tan of 90 degrees minus π is equal to cot π. This means that tan squared of 90
degrees minus π is equal to cot squared π. And the denominator of our
expression is therefore equal to one plus cot squared π.

Letβs now consider the angle three
π over two minus π. Once again, we can see from the
unit circle that three π over two radians is equal to 270 degrees. This means that the numerator of
our expression is equal to one plus cot of 270 degrees minus π. If π lies in the first quadrant,
as shown in our right triangle, then three π over two minus π, or 270 degrees
minus π, lies in the third quadrant. It is clear from the diagram that
cos of three π over two minus π is equal to negative sin π and sin of three π
over two minus π is equal to negative cos π. Since sin π over cos π is tan π
and cos π over sin π is cot π, then cot of 270 degrees minus π is equal to tan
π. Squaring both sides of this
identity, we can rewrite the numerator of our expression as one plus tan squared
π.

Our next step is to recall two of
the Pythagorean identities. Firstly, tan squared π plus one is
equal to sec squared π. And secondly, one plus cot squared
π is equal to csc squared π. Our expression simplifies to sec
squared π over csc squared π. And this can be rewritten as sec
squared π multiplied by one over csc squared π. Recalling the reciprocal identities
sec π is equal to one over cos π and csc π is equal to one over sin π, we have
one over cos squared π multiplied by sin squared π, which can be rewritten as sin
squared π over cos squared π and, in turn, is equal to tan squared π. The expression one plus cot squared
three π over two minus π over one plus tan squared π over two minus π written in
its simplest form is tan squared π.

We will now summarize the key
points from this video. In this video, we simplified
trigonometric expressions using a variety of trigonometric identities. We used the three Pythagorean
identities sin squared π plus cos squared π is equal to one, tan squared π plus
one is equal to sec squared π, and one plus cot squared π is equal to csc squared
π. We also used the reciprocal
identities csc π is equal to one over sin π, sec π is equal to one over cos π,
and cot π is equal to one over tan π. Since tan π is equal to sin π
over cos π, we also saw that cot π is equal to cos π over sin π.

We also recalled the cofunction
identities sin of 90 degrees minus π is equal to cos π and cos of 90 degrees minus
π is equal to sin π. And using the reciprocal identities
above, these led us to four further cofunction identities. Finally, we used the unit circle to
determine other related angle identities as demonstrated in our final example. In many examples, we need to apply
more than one identity or type of identity to simplify a trigonometric
expression.