### Video Transcript

In this video, we will learn how to
simplify a trigonometric expression. First, weβll recap our
trigonometric functions. And then weβll consider reciprocal
trigonometric functions. And then later on, weβll look at
even and odd identities for trigonometric functions. The reciprocal functions and the
even and odd identities will help us to simplify trigonometric expressions.

If we consider this right triangle
with an angle π, we have its opposite side length, its adjacent side length, and
the hypotenuse. The three trigonometric functions
can be expressed in terms of the ratio of the sides of the triangle. sin of π equals the opposite over
the hypotenuse. cos of π equals the adjacent over
the hypotenuse. And tan of π equals the opposite
over the adjacent. And then we have the trigonometric
identity that the tan of π equals the sin of π over the cos of π.

When weβre using trigonometric
functions in right triangles, weβre always dealing with acute angles for π. We can also consider the
trigonometric functions for the values of π defined on the unit circle. For any given point π₯, π¦ that
lies on the unit circle and the angle π, the sine function is defined as π¦ equals
the sin of π and the cosine function is π₯ equals cos of π.

One tool we use to simplify
trigonometric expressions are the reciprocal trigonometric functions. Remember that the reciprocal is
what we multiply a value by to get one. For a number, its reciprocal is
simply one over that number. And here we see that the reciprocal
of sin of π is one over sin π. And that value is called the
cosecant. Therefore, the sec of π equals one
over cos of π. sec π is the reciprocal of cos π. cot of π equals one over tan of
π. By the definition of tangent, we
can rewrite that as the cos π over the sin of π, making the cot π the reciprocal
of tan π.

Now that we know these reciprocal
functions, itβs worth pointing out that when weβre working with trigonometric
expressions, the first step is usually taking all reciprocal functions and rewriting
them in terms of sine and cosine. Letβs see how that would work.

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

In this expression, we have a trig
function and a reciprocal trig function. One strategy for evaluating a
trigonometric expression is to write it in terms of sine and cosine functions. The first term in this expression
is already in terms of sine and cosine. And to rewrite csc π in terms of
sine or cosine, we recall that csc π is the reciprocal of sin π and that csc π
equals one over sin π.

Therefore, in the denominator of
this second term, we replace csc π with one over sin π. If we think about negative five
over one over sin π, this is negative five divided by one over sin π. And dividing by a fraction is the
same as multiplying by the reciprocal of that fraction. Therefore, we can rewrite negative
five over one over sin π as negative five times sin of π. In this case, we have a sin π in
the denominator and a sin π in the numerator, which cancels out, leaving us with
eight times negative five, which is negative 40.

In our next example, weβll simplify
a trigonometric expression without finding a value for that expression.

Simplify cos π csc π sin π.

In this expression, we have two
trigonometric functions and one reciprocal trigonometric function. One strategy for simplifying a
trigonometric expression is to rewrite it in terms of sine and cosine functions. This means taking any reciprocal
functions and rewriting them in terms of sine and cosine. Recall that the csc π is the
reciprocal function of the sine function, meaning it is equal to one over sin
π. This means in our expression, we
can substitute one over sin π in place of csc π. We know that one over sin π times
sin π equals one and cos π times one equals cos π.

Letβs look at one more example of
simplifying trigonometric expressions with reciprocal functions.

Simplify tan π sin π over sec
π.

In this expression, we have
trigonometric functions and a reciprocal trigonometric function. A good strategy for simplifying a
trigonometric expression is to rewrite it in terms of sine and cosine functions. In this case, weβll want to rewrite
the tan of π in terms of sine and cosine and the sec π in terms of sine and
cosine.

Recall that tan π equals sin π
over cos π and that sec π equals one over cos π. In our first step, Iβve rewritten
our expression tan π sin π divided by sec of π, switching the fraction bar for
the division symbol. In the next step, weβll substitute
one over cos π in place of sec π and weβll substitute sin π over cos π in place
of tangent. This means we now have sin π over
cos π times sin π divided by one over cos π. And then we know that dividing by a
fraction is multiplying by its reciprocal, making our new expression sin π over cos
π times sin π times cos π.

If we rewrite sin π over cos π as
sin π times one over cos π, we have a cosine in the denominator and a cosine in
the numerator. When you multiply these together,
they equal one. At this point, weβre left with sin
π times sin π, which equals sin squared π.

Now letβs consider another property
of trigonometric and reciprocal trigonometric functions. Any even function satisfies that π
of negative π equals π of π. And any odd function satisfies that
π of negative π equals the negative π of π. When it comes to trig functions,
the cosine function is even and the sine function is odd. We can see this in the graph of the
cosine function, where the cos of negative 180 degrees equals negative one and the
cos of positive 180 degrees also equals negative one. And with the graph of the sine
function, we see that sin of negative 90 degrees equals negative one and the sin of
90 degrees equals one.

Notice that in the sine function,
the sin of 90 degrees and the sin of negative 90 degrees are opposites of one
another. We can also determine the parody of
the other trigonometric functions that are defined in terms of sine and cosine. For example, the tan of negative π
would be equal to the sin of negative π over the cos of negative π. The sin of negative π equals the
negative sin of π. And the cos of negative π equals
the cos of π. Bringing out the negative, we have
the sin π over cos π, which equals the tan of π. This means the tan of negative π
equals the negative tan of π. And that makes the tangent function
odd.

Therefore, for any value of π in
their respective domains, cosine and secant functions are even. And for any values of π in their
respective domains, sine, tangent, cosecant, and cotangent are odd. This is called the parody of
trigonometric functions. Letβs consider how we might use the
parody to simplify expressions.

Simplify tan of negative π times
csc of π.

In this expression, we have a
trigonometric function of a negative π and a reciprocal function. One strategy for solving
trigonometric expressions is to rewrite the expression in terms of sine and
cosine. Recall that the csc of π equals
one over the sin of π. Therefore, we can rewrite csc π as
one over sin π. Additionally, we recall the parody
of the tangent function, that is, that the tangent function is an odd function. The tan of negative π equals the
negative tan of π. We know that the tan of π equals
sine over cosine. Therefore, the negative tan of π
will be equal to the negative sin π over cos π.

Our new expression is negative sin
of π over cos of π times one over sin of π. The sine in the numerator and the
denominator cancels out. And we have negative one over cos
of π. And we know that sec π equals one
over cos π, which makes this expression simplify to negative sec of π.

Letβs now consider a third set of
identities that help us simplify trigonometric expressions. We could use the graphs of sine and
cosine to explore this identity. If we look on the graph where sin
of π equals one, one of the places this happens is at 90 degrees. If we look at the places where cos
of π equals one, we see that this happens at zero degrees. This illustrates that the sine
function is equivalent to the cosine function by a translation of 90 degrees to the
left. And therefore, we can say that the
sin of 90 degrees plus π equals the cos of π. Additionally, the sin of 90 degrees
minus π will equal the cos of π.

In our example, this means that the
sin of 90 degrees plus zero degrees equals the cos of zero degrees. The output of both of these values
equals one. We can write these two identities
in the form sin of 90 degrees plus or minus π equals cos of π. For cosine, the cos of 90 degrees
plus π will be equal to the negative sin of π. And the cos of 90 degrees minus π
will be equal to the positive sin of π. And the tan of 90 degrees plus or
minus π equals the negative or positive cot of π. Of course, all of this is true if
weβre working in radians. We simply replace 90 degrees with
π over two.

Likewise, we have the three
reciprocal values. cot of 90 degrees plus or minus π equals the negative or
positive tan of π. sec of 90 degrees plus or minus π equals the negative or
positive csc of π. And the csc of 90 degrees plus or
minus π equals the sec of π.

In our next example, weβll use
correlated angle identities in addition to the parody of the trigonometric functions
to simplify an expression.

Simplify sin of π over two plus π
times sec of negative π.

To simplify this expression, recall
that the sec of negative π equals the sec of π. Additionally, the sin of π over
two plus or minus π equals the cos of π based on the correlated angles
identity. Now we have cos π times sec of
π. And finally, remember that the
secant function is the reciprocal of cosine. Replacing sec π with one over cos
π, we have cos π times one over cos π, which equals one.

Weβve already explored some
correlated angle identities that deal with a 90-degree shift. By repeatedly applying the
identities we used earlier in the video or using the unit circle, we could identify
these correlated angle identities that sin of 180 degrees plus or minus π equals
negative or positive sin of π. The cos of 180 degrees plus or
minus π equals the negative cos of π. The tan of 180 degrees plus or
minus π equals the positive or negative tan of π. And of course, if we were working
with the unit circle, this would be π radians. And we can use this to find the
values of the reciprocal functions as well.

Letβs look at another example of
simplifying an expression.

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

In this expression, we have a
reciprocal function divided by a reciprocal function. Additionally, we have a cofunction
identity and a correlated identity. First, the sec of π over two minus
π equals the csc of π. And second, the cot of π minus π
equals the negative cot of π. We rewrite sec of π over two minus
π as csc π. And the cot of π minus π becomes
the negative cotangent. And then weβll recall that our
reciprocal functions csc π equals one over sin π, cot π equals cos π over sin
π.

Weβre using a strategy to take all
of our reciprocal functions and write them in terms of sine and cosine, which gives
us one over sin π divided by negative cos π over sin π. Dividing by a fraction is
multiplying by its reciprocal. A sine in the denominator and a
sine in the numerator cancel each other out, which equals negative one over cos π,
which means weβll need to use one final identity sec π equals one over cos π,
which makes the simplified form negative sec π.

Weβve considered identities for a
90-degree shift and a 180-degree shift of our angle. Now letβs consider the shift of 270
degrees and 360 degrees. The correlated angle identities for
a shift of 270 degrees are as follows. The unit circle allows us to
determine the correlated angle identities for sine and cosine. Letβs consider the right triangle
created at π. Weβll let the length thatβs lined
here in green be π and the length in blue be π. And we know that the hypotenuse is
equal to one, as this is the unit circle. In this case, the cos of π equals
π and the sin of π equals π.

Now, if we consider the angle
created by three π over two minus π, it has a reference angle with the π₯-axis
thatβs highlighted in pink here. Labeling the right triangle created
with the reference angle as π, π, and one and then considering the CAST diagram,
we know that the sine relationship for a point in the third quadrant is going to be
negative, which means the sin of 270 degrees minus π equals negative π. This confirms the identities weβve
listed.

As you can see, you can use the
unit circle to identify all of the correlated angles identities if you donβt
remember them: π over two; π; three π over two; and what weβll list next, two
π. Hereβs the list of the correlated
angle identities for a 270-degree shift and a 360-degree shift. Again, that represents a three π
over two, if youβre working in radians, or two π, respectively.

In the final example, weβll use
these correlated angle identities in order to simplify a trigonometric
expression.

Simplify sec π tan π tan of 270
degrees plus π.

To simplify this expression, letβs
first simplify the tan of 270 degrees plus π using correlated angle identities. Recall that the tan of 270 degrees
plus π equals the negative cot of π. In place of tan of 270 degrees plus
π, we can substitute negative cot of π. We can notice that the tangent and
cotangent are reciprocals of one another. That is, the cot of π equals one
over the tan of π. And when we multiply these values
together, we get negative one. So we have negative one times sec
of π, which means a simplified form of this expression is the negative sec π.

Letβs finish by reviewing the key
points. We can express the tangent and
reciprocal functions in terms of sine and cosine and use these to simplify
trigonometric expressions. These functions are either even or
odd. The cos of negative π equals the
cos of π; itβs even. And the sin of negative π equals
the negative sin of π; itβs odd. The other functions can be derived
from the definition of the sine and cosine functions for even and odd. The unit circle allows us to
determine the correlated angle identities for sine and cosine. And finally, we often need to apply
more than one identity or even more than one type of identity to simplify a
trigonometric expression.