### Video Transcript

In this video, we will learn how to
simplify trigonometric expressions. We will recap the reciprocal
trigonometric functions and recall which functions are odd and which are even. Weβll then look at the Pythagorean
identities and the cofunction identities and learn how to use these to simplify
expressions. Letβs start by recapping some
reciprocal trigonometric functions that we should already know.

csc of π is equal to one over sin
of π, and sec of π is equal to one over the cos of π. We can of course see any of these
functions or identities written with the variable of π₯ instead of π, but the same
rule applies. We also have the function that the
tan of π is equal to sin of π over cos of π and the cot of π is equal to one
over the tan of π. Because the tan of π is equal to
sin of π over cos of π, then the cot of π can also be written as the cos of π
divided by the sin of π.

We can also recall the identities
for negative angles. Because sine and tangent are odd
functions, then the sin of negative π is equal to negative sin π and tan of
negative π is equal to negative tan of π. Cosine is an even function, so the
cos of negative π is equal to the cos of π.

Weβll now look at a few questions
where we use these trigonometric identities to simplify expressions.

Find the value of eight over sin π
multiplied by negative five over csc π.

When we have an expression like
this involving a cosecant, then itβs always useful to remember the reciprocal
identity. csc π is equal to one over sin π. So when we simplify this
expression, weβll replace csc π with one over sin π. When it comes to simplifying this
fraction, negative five over one over sin π, itβs worthwhile remembering that this
is equivalent to negative five divided by one over sin π. And to divide by a fraction, we
multiply by its reciprocal. So weβll be working out negative
five multiplied by sin π over one, which gives us negative five sin π.

So now when we return to the
original calculation, weβll be working out eight over sin π multiplied by negative
five sin π. Considering the second value as a
fraction over one, we then realize that we can divide through by sin of π. And so weβre left with eight
multiplied by negative five divided by one, which gives us negative 40 over one or
just simply negative 40. Therefore, weβve answered the
question. The value of eight over sin π
multiplied by negative five over csc π is equal to negative 40.

Letβs take a look at another
question.

Simplify cos squared π sec π csc
π.

When we have questions like this
that involve a large number of functions, it can initially be quite confusing. However, the best thing to do is
think about the different functions that make up our expression and see if we can
recall any identities regarding these. As we have sec of π and csc of π,
it might be useful to recall that the sec of π is equal to one over the cos of π
and the csc of π is equal to one over sin of π. We can then replace these two
functions with their reciprocal functions, which gives us cos squared π multiplied
by one over cos of π multiplied by one over sin π.

When we simplify this expression,
we have cos squared π over cos π sin π. At this point, it might be worth
reminding ourselves that cos squared π is just the same as cos π multiplied by cos
π. Now, when we look at this
expression, we should notice that we can take out a common factor of cos π from the
numerator and denominator. This leaves us with cos π over sin
π, and this expression should look familiar. The cot of π is in fact equal to
cos of π over sin of π. And so what we have here is the cot
of π. And so our answer is that cos
squared π sec π csc π simplifies to cot π.

In the next question, weβll need to
use some of the even and odd identities to help us simplify a trigonometric
expression.

Simplify tan of negative π csc of
π.

When weβre simplifying
trigonometric expressions, if we have a look at the question, it will give us a clue
as to which identity weβll need to use. The angle of negative π here
should make it clear that we probably need to use the even and odd identities or
often called the negative angle identities. The tangent function is an odd
function. So that means that tan of negative
π is equal to negative tan of π. As this function also involves the
csc of π, itβs worthwhile remembering the reciprocal identity. The csc of π is equal to one over
sin of π.

We can then plug in these
values. The tan of negative π is equal to
negative tan of π, and csc of π is equal to one over sin of π. Sometimes when weβre simplifying
trigonometric expressions, the best thing we can do is to see if we can write our
expressions in terms of sine or cosine. If we look at this negative tan of
π, we can in fact replace this with something else. The tan of π is equal to sin π
over cos π.

We can then plug this value into
our expression for the tan of π, being careful that we donβt lose this negative
sign along the way. Before we simplify negative sin π
over cos of π multiplied by one over sin π, observe that we can take out a common
factor of sin of π from the numerator and denominator. Multiplying what remains then, we
have negative one over cos of π.

We should remember that sec of π
is equal to one over cos of π. But as we have negative one over
cos of π, the answer will be negative sec of π. And so weβve simplified tan of
negative π csc of π to negative sec of π.

Before we look at any more
questions, letβs have a look at the Pythagorean identity for trigonometry. Itβs perhaps the most commonly used
of all the trigonometric identities. And it tells us that sin squared π
plus cos squared π is equal to one. We can of course rearrange this
into different formats. For example, by subtracting cos
squared π from both sides, we would find that sin squared of π is equal to one
minus cos squared of π. We can even manipulate it by
dividing through by cos squared π.

Simplifying the first term of sin
squared π over cos squared π gives us tan squared π. The next term of cos squared π
over cos squared π would simplify to one. On the right-hand side, one over
cos squared π simplifies to sec squared π. And we know this because one over
cos π is equal to sec π. Knowing both of these identities
off by heart will allow us to very quickly manipulate and simplify trigonometric
expressions.

Weβll now have a look at some
questions where weβll use these identities along with the other ones that we already
know.

Simplify sin π plus cos π squared
minus two sin π cos π.

When we first look at this
question, it might seem a little more complicated because we donβt know which
identities weβll need to use. However, letβs see if we can do
anything with this sin π plus cos π squared. When we square a value, it just
means weβre multiplying it by itself. We can then expand these
parentheses using a method like the FOIL method. sin of π multiplied by sin of π
gives us sin squared π. We have sin of π multiplied by cos
of π two times. And finally cos π multiplied by
cos π gives us cos squared π.

An alternative to multiplying out
the double brackets in this way is to remember that if we have this bracket
multiplied, we square the first term, we add on the square of the second term, and
then we add on double the product of both terms. Either method would give us sin
squared π plus two sin π cos π plus cos squared π minus two sin π cos π. When we look at this expression, we
can observe we have two sin π cos π minus two sin π cos π, which is equivalent
to zero.

Now, all that weβre left with is
sin squared π plus cos squared π. And this expression should be
familiar. The Pythagorean identity tells us
that sin squared π plus cos squared π is equal to one. This is exactly the same as weβve
found in our workings. And so we can give the answer that
sin π plus cos π squared minus two sin π cos π simplifies to one.

Letβs have a look at another
question involving this Pythagorean identity.

Simplify sin π csc π minus cos
squared π.

When we look at a question like
this, perhaps the most obvious place to start would be to see if we can rewrite csc
of π in a different way. We can remember that the csc of π
is equal to one over sin π. When we replace csc of π with one
over sin π, we end up with the expression sin of π multiplied by one over sin of
π minus cos squared π. Working out sin π multiplied by
one over sin π will give us sin of π over sin of π. And this of course simplifies to
one. So how do we simplify one minus cos
squared π any further?

Well, we can remember the
Pythagorean identity, which tells us that sin squared π plus cos squared π is
equal to one. Rearranging this equation by
subtracting cos squared π from both sides would give us that sin squared π is
equal to one minus cos squared π. And this is exactly what we have in
our simplified expression. One minus cos squared π is equal
to sin squared π. And so weβve got the answer that
sin π csc π minus cos squared π can be simplified to sin squared π.

Weβll now look at a different set
of trigonometric identities, the cofunction identities, which is when we express
trigonometric functions in terms of their compliments. So we have cos of π equals sin of
π over two minus π, sin of π equals cos of π over two minus π, cot of π equals
tan of π over two minus π, tan of π equals cot of π over two minus π, the csc
of π equals sec of π over two minus π, and finally sec π equals csc of π over
two minus π.

Itβs worth pointing out here that
all of these angles will be expressed in radians. But we could also express them in
terms of degrees. For example, the cos of π is equal
to sin of 90 degrees minus π. It may be worth pausing the video
to make a note of these identities as weβll be using some of them in the final
question.

Simplify sec of π over two minus
π over cot of π minus π.

When we have a trigonometric
expression which involves an angle of π over two minus π, itβs a good indication
that weβll probably be using the cofunction identities. The cofunction identity for sec of
π over two minus π tells us that this is equivalent to csc of π. Therefore, when we simplify this
expression, the numerator will be csc of π. So what about this denominator?

Well, one possible way to simplify
this is to remember that the cot of π is equal to cos π over sin of π. So for our value of cot π minus
π, this will become cos of π minus π over sin of π minus π.

Next, letβs use the fact that cos
of π minus π equals negative cos π and sin of π minus π equals sin π. Therefore, our expression has csc
π on the numerator and negative cos π over sin π on the denominator. Before we tidy up this fraction on
the denominator, letβs see if we can do anything more with the csc of π on the
numerator.

We should remember one of the
reciprocal identities, which tells us that the csc of π is equal to one over sin of
π. Therefore, on the numerator, we can
replace csc of π with one over sin π. This expression now does look very
confusing with a fraction on the numerator and denominator. It can be helpful to remember that
a fraction simply means the numerator divided by the denominator.

Remember that when weβre dividing
fractions, we simply multiply by the reciprocal of the second fraction. We notice that before we multiply,
we can take out this common factor of sin π from the numerator and denominator. Multiplying these values then gives
us one over negative cos π. We can use one final trigonometric
identity. The sec of π is equal to one over
the cos of π. As we have one over negative cos
π, then our value will simplify to negative sec π. And so weβve answered the question
to simplify sec of π over two minus π over cot of π minus π.

We can now recap some of the
identities that we saw in this video. We began this video by recapping
some of the reciprocal identities. We then reminded ourselves of the
negative angle or odd and even function identities. We looked at the Pythagorean
identity sin squared π plus cos squared π equals one. And we also saw how we can use this
to derive sec squared π equals one plus tan squared π. Finally, we also saw the cofunction
identities. And finally, itβs worth remembering
that we often need to apply more than one identity or even type of identity to
simplify a trigonometric expression.