### Video Transcript

Consider the following arc of a
unit circle, where ray ππ is inclined at π radians. Part one) What, in terms of π, are
the coordinates of π?

Well, here is π. And if we take the angle πππ to
be a right angle, we can see that π is directly above the point π with coordinates
one, zero. And so, the π₯-coordinate of π
must be one. What about the π¦-coordinate
though? Well, the π¦-coordinate of π is
just the length ππ. We find this length by considering
the right triangle πππ. The origin π has coordinates zero,
zero. And so, we can see that the length
ππ is just one.

So we know the length of the side
adjacent to the angle π. And weβd like to know the length of
the side opposite it. Well then, we should use tan π,
which is the ratio of the length of the side opposite the angle π. Thatβs ππ divided by the length
of the side adjacent which is side ππ, which we know has length one. ππ which remember is the
π¦-coordinate of π is, therefore, tan π. And so, the coordinates of π are
one, tan π.

Part two) Write the following
inequalities in terms of sin π, π, and cos π. The length of ππ
is less than the
length of the arc from π to π, which is less than the length ππ.

Letβs start with ππ
. Again, itβs best to find this by
considering a right triangle, in this case the right triangle ππ
π. Now, as π lies on the unit circle,
its distance from the origin is one. And so, ππ is one. We have the length of the
hypotenuse of the right triangle. And we want to find the length of
the side opposite to the angle π. Thatβs the length ππ
. So as weβd like to find the
opposite side and we have the hypotenuse, we use sine. Sin of π is opposite β thatβs ππ
β over hypotenuse one. And so, ππ
is sin π.

We now find the length of the arc
from π to π. Thatβs this length here. Well, we can see that this arc
subtends an angle of π at the centre of this unit circle. And π weβre told in the question
is measured in radians. By definition then, the length of
the arc from π to π is π. Alternatively, you manage to find
π in radians as the ratio of the arc length to the radius. But as weβre in the unit circle,
the radius is one. And so, π is just the arc length
ππ. What about the length ππ?

Well, if you remember in the first
part of the question, we found this to be tan π. ππ was the π¦-coordinate of the
point π. But weβre asked to give our answer
in terms of sin π, π, and cos π alone. So we rewrite tan π as sin π over
cos π. Our inequality becomes that sin π
is less than π which is less than sin π over cos π.

Finally, part three) by dividing
your inequalities by sin π, using the squeeze theorem and the fact that the limit
of cos π as π approaches zero is one, which of the following conclusions can you
draw? Is it that A) the limit of π over
sin π as π approaches zero is zero, B) that this limit does not exist, or C) that
this limit has value one.

We start by dividing the
inequalities obtained in the second part of the question by sin π. Sin π divided by sin π is
one. π divided by sin π canβt be
simplified further. But sin π over cos π divided by
sin π can. We get one over cos π. Now, we use the squeeze
theorem. But how should we use it? Well, the options give us some
clue. Basically, weβre asked for the
limits of π over sin π as π approaches zero. So what is this limit? Well, weβve got two functions
either side of π over sin π. The limit of one is of course just
one. The limit of one over cos π is
slightly more tricky. We have to use the fact that the
limit of a quotient is the quotient of the limit.

But having done this, we know what
the limit of cos π as π approaches zero is. Weβre told that in the
question. This value is one. And so, the limit of one over cos
π as π approaches zero is one over one which is one. So both the function one and the
function one over cos π have the same limit: the limit one as π approaches
zero. And so, as π over sin π lies
between these two functions for values of π near to zero, we can apply the squeeze
theorem. The limit of π over sin π as π
approaches zero must be one as well, the same as the other two limits. This is our answer which
corresponds to option C.