### Video Transcript

Suppose π is a point on the
unit circle corresponding to the angle four π over three. Is there another point on the
unit circle representing an angle in the interval zero to two π that has the
same tangent value? If yes, give the angle.

First, we might wanna sketch a
coordinate plane and then add a unit circle, which is a circle with the center
at the origin and a radius of one. From there, we might also want
to label our coordinate plane with radians beginning at zero, π over two, π,
three π over two, and two π. Because we have the interval
zero to two π, we know weβre only interested in one full turn. Our point π is on the unit
circle and corresponds to the angle four π over three. This means our first job is to
find out where the angle four π over three would land.

I know for π over three is
greater than π, but itβs probably worth comparing to find out if four π over
three is greater than or less than three π over two. If we give these fractions
common denominators, four π over two becomes eight π over six and three π
over two becomes nine π over six. Since eight π over six is less
than nine π over six, we can say that four π over three is less than three π
over two. And that means point π is
going to fall in our third quadrant and that four π over three would be this
angle.

Since we know that our angle
falls in the third quadrant, we can use the CAST diagram, which will tell us
that the tangent of the angle in the third quadrant is going to be positive. In order for us to find another
point inside the unit circle that has the same tangent value, weβll be looking
for the other place where the tangent value could be positive. And that will be in the first
quadrant. In the first quadrant, all trig
values will be positive.

But in order for us to find
what the value of that angle would be in the first quadrant, we need to break up
our angle four π over three into smaller parts. We could say that four π over
three is equal to π plus π over three, the distance from zero to π and then
an additional π over three. The right triangle created
inside the unit circle four π over three in the third quadrant would look like
this.

And in the first quadrant,
there would be some point such that we would be dealing with the angle of π
over three. In the first quadrant, that
would have π₯, π¦ coordinates. And in the third quadrant, it
would have negative π₯, negative π¦ coordinates. And we know that in a unit
circle, the tan of π will be equal to π¦ over π₯. And we would say that the tan
of four-thirds π is equal to negative π¦ over negative π₯ and the tan of π
over three is equal to π¦ over π₯. But we simplify negative π¦
over negative π₯ to just π¦ π₯. And so, weβve shown that yes,
there is another angle in this interval that has the same tangent value. And itβs the angle π over
three.