### Video Transcript

In this video, we will learn how to
use cofunction and even odd identities to find the values of trigonometric
functions. Trigonometric functions have many
different properties and identities that help us simplify and solve equations. For this lesson, we want to review
cofunction identities, even odd identities, and then use them to solve some
problems.

First, letβs consider the
cofunction identities, which tell us sin of 90 degrees minus π equals the cos of
π. cos of 90 degrees minus π equals
the sin of π. And the tan of 90 degrees minus π
equals the cot of π. We should also remember that if
weβre working in radians, weβll substitute π over two in place of 90 degrees.

We can show these cofunction
identities graphically using the unit circle. If we have an angle π in the unit
circle, we can say that it forms a right triangle with a hypotenuse of one and side
lengths π and π. We know the sin of π will be equal
to the opposite over the hypotenuse, which in this case would be π over one. So sin of π equals π. The cos of π is the adjacent side
length over the hypotenuse, π over one, which is just π. And the tan of π is the opposite
over the adjacent side length, π over π. And we want to explore a
relationship between π and 90 minus π.

In the unit circle on the
coordinate grid, the space between our angle and 90 degrees can be written as 90
minus π. And so we can create a second right
triangle in quadrant one, which will again have a hypotenuse of one and side lengths
of π and π. Looking at our new triangle in
yellow, if we take the sin of 90 minus π, thatβs the opposite over the hypotenuse,
which in this case will be π over one. Therefore, sin of 90 minus π
equals π. And the cos of 90 minus π is the
adjacent over the hypotenuse. In this case, thatβs π. And the tan of 90 degrees minus π
will then be π over π, and so weβve demonstrated these cofunction pairs. sin of π and cos of 90 degrees
minus π both equal π here, which is what we should expect based on our cofunction
identity.

Weβve also demonstrated that both
cos of π and sin of 90 degrees minus π equal π. That is to say that sin of 90
degrees minus π is equal to cos of π. The tangent pair, the tangent
cofunction identity, is a little bit different. Notice that the tan of 90 degrees
minus π equals π over π. But the tan of π equals π over
π. If we take the reciprocal of the
tan of π, that is, one over tan of π, it would be equal to π over π. And then we recognize that the
reciprocal of the tan of π is equal to the cot of π as the cot of π is equal to
the adjacent side length over the opposite side length, which demonstrates that
third cofunction identity that the tan of 90 minus π is equal to the cot of π.

Letβs now consider the odd and even
trigonometric function identities. The odd and even identities are as
follows. sin of negative π is equal to
negative sin of π, which makes sine an odd function. cos of negative π is equal to cos
of π, which makes cosine an even function. And the tan of negative π is equal
to the negative tan of π, which makes tangent an odd function. Again, we can demonstrate this with
the unit circle. For an angle π, in the first
quadrant of the unit circle, we can create a right-angled triangle with side lengths
π and π, and then there would be some angle negative π. Remember this negative tells us
weβre moving clockwise instead of counterclockwise, making our angle fall in
quadrant four. And again weβll see that this angle
can create a right triangle with side lengths π and π and the hypotenuse of
one.

In order to find the sin of
negative π, weβll use our CAST diagram which helps us identify the signs of the
trigonometric ratios in the four quadrants. In quadrant four, only the cosine
function will be positive. The sign in tangent functions will
be negative. Since the sine relationship is the
opposite side length over the hypotenuse, in quadrant four, this will be negative
π. The cosine relationship is the
adjacent side length over the hypotenuse, which is π over one. And in quadrant four, the cosine
relationship is positive. So the cos of negative π equals
positive π. Similarly, the tan of negative π
will be the opposite over the adjacent side length and in quadrant four will be
negative, which gives us the tan of negative π is negative π over π.

If we want to find the sin of π,
in our diagram, thatβs in quadrant one, it will be equal to π over one. The cos of π is equal to π over
one, and the tan of π is equal to π over π. Notice that the cosine
relationships the cos of negative π is equal to π and the cos of positive π is
equal to π. We recognize this as our even
function and our even function identity. But what about the other two
relationships, the odd functions? If we multiply sin of π and π by
negative one, we can say that the negative sin of π equals negative π. Similarly, if we multiply tan by
negative one, we get negative tan of π. And then we need to multiply the
other side of the equation by negative one, which will give us negative π over
π. And so we would have to say that
sin of negative π is equal to negative sin of π and that the tan of negative π is
equal to the negative tan of π.

Letβs now look at an example where
we need to use these identities to help us solve a trigonometric equation.

Find the value of cos of 90 degrees
plus π given the sin of π equals three-fifths, where π is between zero and 90
degrees.

Weβre given the sin of the angle of
π, and we need to find the cos of 90 degrees plus π. We could do this with a graph. However, a much more
straightforward approach is by rewriting this expression using cofunction
identities. In fact, using cofunction
identities, thereβs a few different ways that we could rewrite this. Weβll consider one of those
ways. If we want to use this cofunction
identity on the cos of 90 degrees plus π, we can rewrite that as the cos of 90
degrees minus negative π. And then using our cofunction
identity, we can say that the cos of 90 degrees minus negative π must be equal to
the sin of negative π.

Thinking about the sin of negative
π, we recognize that the sine function is an odd function. And the sin of negative π is equal
to the negative sin of π. And we know what the sin of π is
equal to. The sin of π in this case is
three-fifths, which means the sin of negative π equals negative three-fifths. We began with cos of 90 degrees
plus π. And using cofunction identities and
odd function identities, we were able to find that it will be equal to negative
three-fifths. Before we move on, we can show one
graphical representation of this. If we draw a unit circle with a
measure of angle π, we know that π falls between zero and 90 degrees. And if the sin of π equals
three-fifths, then the opposite over the hypotenuse must be equal to
three-fifths.

We want to know the cos of 90
degrees plus π. If we add 90 degrees to this angle,
we have a radius of one because this is the unit circle. And because weβve rotated this 90
degrees counterclockwise, the distance from our point to the π¦-axis is
three-fifths. And when weβre working with the
unit circle, the cos of our angle will be equal to the π₯-coordinate, which in our
case will be negative three-fifths, and confirms that the cos of 90 degrees plus π
is equal to negative three-fifths.

In our next example, weβre
combining multiple trig functions. This makes the graphical approach
much more difficult, so weβll need to use a combination of different cofunction
identities to simplify.

Find the value of sin of 180
degrees minus π₯ plus tan of 360 degrees minus π₯ plus seven times sin of 270
degrees minus π₯ given that sin of π₯ equals three-fifths, where π₯ is between zero
and 90 degrees.

We have an expression we want to
simplify that has three terms. And we know that sin of π₯ equals
three-fifths. To simplify the whole expression,
weβll consider each of these terms in turn, beginning with sin of 180 degrees minus
π₯. One of our cofunction identities
tells us that the sin of 90 degrees minus π equals cos of π. Now, our sine term doesnβt look
exactly like this, but we could rewrite it to be sin of 90 degrees plus 90 degrees
minus π₯. We want something in the form 90
degrees minus π. And that means we can rearrange
this by calling it 90 degrees minus π₯ minus 90 degrees, where weβll let π be equal
to π₯ minus 90 degrees. So cos would be equal to π, π₯
minus 90 degrees. This looks very close to the cos of
90 degrees minus π. But in order to rearrange this part
of the function, we need to consider the fact that the cosine function is an even
function.

We know that the cos of negative π
is equal to the cos of π. This means the cos of π₯ minus 90
will be equal to the cos of the negative of π₯ minus 90, which is the cos of 90
degrees minus π₯. The cos of 90 degrees minus π
equals the sin of π. And so we can say that the sin of
180 degrees minus π₯ is equal to the sin of π₯. That simplifies our first term. For our second term, instead of
using cofunction identities, weβll recall the periodic property of the tangent,
which tells us tan of π plus or minus 180 degrees equals the tan of π. We want to rewrite this in a form
thatβs usable to us, so we can say itβs tan of 180 degrees plus 180 degrees minus
π₯. If we let π be equal to 180
degrees minus π₯, then weβre now adding π plus 180 degrees inside this tangent,
which will be equal to tan of 180 degrees minus π₯.

We want to use this periodic
property a second time, so weβll let π be equal to negative π₯. Weβre saying the tan of 180 degrees
minus π₯ is equal to the tan of negative π₯ plus 180 degrees. And that periodic property lets us
simplify that to tan of negative π₯. Because we know the tangent
function is odd, so the tan of negative π₯ will be equal to the negative tan of
π₯. Our expression now is the sin of π₯
minus the tan of π₯. And we need to simplify this third
term. This time, weβll use a periodic
property of sin that the sin of π plus or minus 360 degrees equals sin of π. If we let π be equal to 270 minus
π₯, we can subtract 360 degrees from the argument inside. 270 minus 360 is negative 90. The argument inside the sine
becomes negative 90 degrees minus π₯. Because sine is an odd function,
seven times sin of negative 90 degrees minus π₯ is equal to negative seven sin of 90
degrees plus π₯.

To use our cofunction identity
here, weβll rewrite that to be 90 minus negative π₯, which simplifies to the
negative seven cos of negative π₯. And since cosine is an even
function, negative seven times cos of negative π₯ will be equal to negative seven
times cos of π₯. Our new expression is then sin of
π₯ minus tan of π₯ minus seven times cos of π₯. And since we know that π₯ is an
acute angle, we can solve for the other values using right-triangle
trigonometry. Sketching a right triangle with an
angle of π₯, we know that sin is three-fifths. That is the opposite over the
hypotenuse. At this point, we recognize that
this is a three-four-five right triangle.

Of course, we could use the
Pythagorean theorem to solve for that missing angle. But because it falls in the ratio
with the sine relationship of three to five, we know that that third side must be
equal to four. sin of π₯ equals three-fifths minus
the tan of π₯ β which will be the opposite over the adjacent, three-fourths β minus
seven times the cos, which will be four over five, the adjacent side over the
hypotenuse. And three-fifths minus
three-fourths minus seven times four-fifths will be equal to negative twenty-three
fourths.

Letβs consider one final example
where weβll use cofunction identities to help us evaluate some relationships inside
a triangle.

π΄π΅πΆ is a right triangle at
π΅. Find the cot of πΌ given that the
cot of π is four-thirds.

Because we know that π΄π΅πΆ is a
right triangle, we can also say that π΄π΅π· is a right triangle. And this means we can identify
angle π΄π·π΅ as 90 degrees minus π. It also means we can say that πΌ
plus 90 degrees minus π is equal to 180 degrees as we know that π΅πΆ forms a
straight line. And then if we subtract 90 degrees
from both sides of this equation, we see that 90 degrees equals πΌ minus π. And adding π to both sides tells
us that πΌ equals 90 degrees plus π, which means the cot of πΌ is equal to the cot
of 90 degrees plus π. And now it seems like weβre getting
closer because we know cot in terms of π. Based on our cofunction identity,
we know that the tan of 90 degrees minus π equals the cot of π, so we want to
rearrange the cot of 90 degrees plus π.

We can rewrite it to be the cot of
90 degrees minus negative π. And then, weβll rewrite this in
terms of tangent because the cotangent is the reciprocal of tangent. We can say that this is equal to
one over tan of 90 degrees minus negative π and that tan of 90 degrees minus
negative π simplifies to the cot of negative π. But now we have one over cot of
negative π, which will be equal to the tan of negative π. And since tan of negative π is
equal to the negative tan of π, itβs an odd function, which means we simplify to
the negative tan of π.

Going back to our diagram, if the
cot of π is four-thirds, π΄π΅ equals four, π΅π· equals three, the tan of π is the
opposite over the adjacent side length, which here is three-fourths. And we need the negative tangent,
which will be negative three-fourths. Weβve shown that the cot of πΌ will
be equal to the negative tan of π and itβs negative three-fourths.

Before we finish, letβs quickly
review a few key points. Cofunction identities and even odd
identities facilitate simplification of trigonometric expressions. These identities can be combined
with other trigonometric identities and properties to help us evaluate
expressions. Here are the three cofunction
identities we explored and the three even odd identities we explored.