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 consider three types of
trigonometric identity, the Pythagorean identities, the reciprocal identities, and
the cofunction identities.
Letβs begin by considering the
properties of the unit circle. We recall that the unit circle is a
circle with radius one as shown. It allows us to measure the sin,
cos, and tan of any angle π, as the π₯-coordinate of any point on the circumference
represents cos π and the π¦-coordinate is sin π. The right triangle in our diagram,
together with the Pythagorean theorem, leads us to our first Pythagorean
identity. sin squared π plus cos squared π
equals one. Recalling the reciprocal
trigonometric identities enables us to form two further Pythagorean identities.
The three reciprocal functions are
the secant, cosecant, and cotangent, where the sec of angle π is equal to one over
the cos of π. The csc of π is equal to one over
sin π, and the cot of π is equal to one over tan π. It is also worth noting that since
sin π over cos π is equal to tan π, then cos π over sin π is cot π. The next Pythagorean identity is
found by dividing both sides of our first identity by cos squared π. sin squared π divided by cos
squared π is tan squared π. cos squared π divided by cos
squared π is equal to one. And finally, one divided by cos
squared π is equal to sec squared π. Our second identity states that tan
squared π plus one is equal to sec squared π.
Returning to our first identity,
weβll now divide each term by sin squared π. This gives us one plus cot squared
π is equal to csc squared π. We now have a set of three
Pythagorean identities. We will now look at a couple of
examples where we need to simplify trigonometric expressions.
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 we should do first. However, as a general rule, it is
worth starting by replacing any reciprocal functions with the sine, cosine, and
tangent functions. 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. So we are left with one minus cos
squared π.
Next, we recall one of our
Pythagorean identities. sin squared π plus cos squared π
is equal to one. Subtracting cos squared π from
both sides, we have sin squared π is equal to one minus cos squared π. This means that our expression
simplifies to sin squared π. sin π multiplied by csc π minus
cos squared π in its simplest form is sin squared π.
We will now look at 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, we have the second identity. One plus cot squared π is equal to
csc squared π. We know this based on the
reciprocal trig functions where cos π over sin π is cot π and one over sin π is
csc π. The numerator of our expression is
the same as the left-hand side of the first identity. We can therefore replace sin
squared π plus cos squared π with one. 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 here is the
same as the denominator of our expression. This means that this is also equal
to one. The expression simplifies to one
divided by one. We can therefore conclude that sin
squared π plus cos squared π divided by csc squared π minus cot squared π is
equal to one.
Before looking at our next example,
we will recall the cofunction identities. Once again, we consider the unit
circle we saw earlier. 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 of our triangle is 90 degrees minus π. The coordinates of the point marked
on the unit circle will be the 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 is equal to sin π. Likewise, the sin of 90 degrees
minus π is equal to the cos of π. Since sin π over cos π is equal
to tan π, the tangent of 90 degrees minus π is equal to cos π divided by sin
π. And using our knowledge of the
reciprocal functions, this is equal to the cot of π. It follows that the sec of 90
degrees minus π is equal to csc π. The csc of 90 degrees minus π is
the sec of π. And the cot of 90 degrees minus π
is the tan of π.
Using our knowledge of
complementary angles, we have these six equations known as the cofunction
identities. It is important to note that if our
angles are given in radians, the π over two radians is equal to 90 degrees. We will now look at an example
where we need to use these identities.
Simplify cos π multiplied by csc
of 90 degrees minus π minus tan π multiplied by tan of 90 degrees minus π.
When answering a question of this
type, our first step is to try and simplify our expression using the cofunction
identities. Using our knowledge of
complementary angles and the unit circle, we know that the sin of 90 degrees minus
π is equal to cos π and the cos of 90 degrees minus π is equal to sin π. The cosecant function is the
reciprocal of the sine function. It follows that the csc of 90
degrees minus π is equal to one over cos π, which is equal to sec π. We can therefore rewrite the first
term of our expression as cos π multiplied by sec π.
Next, recalling that sin π over
cos π is tan π, then the tan of 90 degrees minus π is equal to cos π over sin
π, which is equal to the cot of π. The second term of our expression
simplifies to tan π multiplied by cot π.
Our next step is to use our
knowledge of the reciprocal functions. We know that sec π is equal to one
over cos π and cot π is equal to one over tan π. Our expression becomes cos π
multiplied by one over cos π minus tan π multiplied by one over tan π. Both parts of our expression
simplify to one. So we are left with one minus
one. We can therefore conclude that cos
π multiplied by csc of 90 degrees minus π minus tan π multiplied by tan of 90
degrees minus π is equal to zero.
In our final example, we will look
at a more complicated problem and weβll use a variety of the identities seen in this
video.
Simplify one plus cot squared three
π over two minus π divided by one plus tan squared three π over two minus π.
In order to answer this question,
we will need to use a variety of trigonometric identities. In problems of this type, it is not
always clear where we should start. And in this question, this is
complicated further by the angle three π over two minus π. As a result, we will begin by
letting πΌ equal three π over two minus π. This allows us to rewrite our
expression as one plus cot squared πΌ divided by one plus tan squared πΌ. The Pythagorean identities state
that sin squared πΌ plus cos squared πΌ is equal to one. tan squared πΌ plus one is equal to
sec squared πΌ. And one plus cot squared πΌ is
equal to csc squared πΌ.
We notice that the numerator of our
fraction is the same as the left-hand side of the third identity. This means we can rewrite this as
csc squared πΌ. The denominator of our expression
is the same as the left-hand side of the second identity. We can therefore rewrite the
expression as csc squared πΌ over sin squared πΌ. Two of the reciprocal trigonometric
identities state that csc πΌ is equal to one over sin πΌ and sec πΌ is equal to one
over cos πΌ. The second identity could also be
rewritten as one over sec πΌ is equal to cos πΌ.
Rewriting csc squared πΌ as one
over sin squared πΌ and one over sec squared πΌ as cos squared πΌ, we have one over
sin squared πΌ multiplied by cos squared πΌ. This is equal to cos squared πΌ
over sin squared πΌ, which in turn is equal to cot squared πΌ. Replacing πΌ with three π over two
minus π, we have cot squared of three π over two minus π. We might think this is our final
answer. However, we can simplify this one
stage further. And this can be done by considering
related angles in the unit circle. We know that the point shown on the
unit circle in the first quadrant has coordinates cos π, sin π. We know that three π over two
radians is equal to 270 degrees. The point with coordinates cos of
three π over two minus π, sin of three π over two minus π is as shown.
We notice that the displacement in
the negative π₯-direction of this triangle is the same as the displacement in the
positive π¦-direction of our first triangle. This means that the cos of three π
over two minus π is equal to the negative of sin π. Likewise, the sin of three π over
two minus π is equal to negative cos π. Since cos π divided by sin π is
equal to cot π, dividing these two equations we have cot of three π over two minus
π is equal to negative sin π over negative cos π. The right-hand side simplifies to
tan π. cot squared three π over two minus π is therefore equal to tan squared
π. The initial expression one plus cot
squared three π over two minus π divided by one plus tan squared three π over two
minus π 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. Firstly, the Pythagorean
identities. sin squared π plus cos squared π
equals one, tan squared π plus one equals sec squared π, and one plus cot squared
π is equal to csc squared π. We also use 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 sin π over cos π is tan π,
we also saw that cos π over sin π is cot π.
Finally, we saw the cofunction
identities. The sin of 90 degrees minus π is
equal to cos π and the cos of 90 degrees minus π is sin π. Using the reciprocal identities,
weβre also able to form four further cofunction identities. csc of 90 degrees minus
π is sec π, sec of 90 degrees minus π is csc π, tan of 90 degrees minus π is
cot π, and finally cot of 90 degrees minus π is tan π.