### Video Transcript

A particle following the
parameterization π₯ is equal to the cos of two π multiplied by π‘ and π¦ is equal
to the sin of two π multiplied by π‘ of the unit circle starts at the point one
zero and moves counterclockwise. At what values of π‘ is between
zero and four inclusive is the particle at the point zero, one? Give exact values.

The question gives us a pair of
parametric equations for the position of the particle. This means that our first equation
π₯ is equal to the cosine of two π multiplied by π‘ gives us the π₯-coordinate of
the position of the particle at π‘. And our second equation of π¦ is
equal to the sin of two π multiplied by π‘ gives us the π¦-coordinate of the
position of the particle at π‘. So finding the values of π‘ where a
particle is at the point zero, one is equivalent to finding the values of π‘ at
which the π₯-coordinate is equal to zero and the π¦-coordinate is equal to one.

So if the π₯-coordinate of the
position of the particle is zero, this is the same as saying that zero is equal to
the cos of two π multiplied by π‘. And if the π¦-coordinate of the
position of the particle is equal to one, this is the same as saying that one is
equal to the sin of two π multiplied by π‘. Where we must be careful because
the question restricts the values of π‘ to be between zero and four inclusive. So we have to solve these two
equations for values of π‘ between zero and four inclusive.

There are a multiple different ways
of solving equations like this. Weβre going to use a sketch. So we start by sketching a graph of
the cos of π₯. We want to know where this will be
equal to zero. Since this will help us determine
where the cosine of two π multiplied by π‘ will be equal to zero. We see that our graph of the cos of
π₯ is equal to zero when π₯ is equal to π over two, three π over two, five π over
two, seven π over two, and so on. We can see that these are just
multiples of π added to or subtracted from π over two. Or, more formally, π over two plus
π multiplied by π, where π is an integer.

So what we have shown is that if
the cos of π₯ is equal to zero, then π₯ is an integer multiple of π plus π over
two. We can then replace the π₯ in this
statement with two π multiplied by π‘. To give us that if the cos of two
π multiplied by π‘ is equal to zero, then two π multiplied by π‘ is equal to π
over two plus an integer multiple of π. We can then solve this by dividing
both sides of the equation by two π, giving us that π‘ is equal to a quarter plus
π divided by two, where π is any integer. Since the question tells us that π‘
is to be restricted between zero and four inclusive, letβs try substituting in π as
the positive integers.

So we get that π‘ is equal to a
quarter, a quarter plus a half, a quarter plus two over two, etcetera. If we then calculate these values
for π‘, we get that π‘ is equal to a quarter, three-quarters, five-quarters. And we carry this on all the way up
to 15 divided by four where we stop because we know that π‘ must be less than or
equal to four. So what we have just shown is that
these are the values of π‘ between zero and four inclusive where the cos of two π
multiplied by π‘ is equal to zero. And so, the π₯-coordinate of the
position of our particle is equal to zero.

We can now do the same to find the
values of π‘ where the π¦-coordinate of the position of our particle is equal to
one. So we sketch the graph of the sign
of π₯. We see that the graph of the sign
of π₯ is equal to one when π₯ is equal to π over two or five π over two, and so
on. A fact we know about the sine
function is that the sin of π₯ is equal to the sin of π₯ plus or minus two π. So in particular, the sin of π₯ is
equal to one when π₯ is equal to π over two plus an even integer multiple of
π. Therefore, what we have shown is
that sin of two π multiplied by π‘ is equal to one whenever two π multiplied by π‘
is equal to π over two plus an even integer multiple of π.

Next, we can divide both sides of
this equation by two π to give us an expression for π‘. Which we can calculate to give us
that π‘ is equal to a quarter plus any integer π. Just as we did before, we want the
values of π which give us a π‘ between zero and four inclusive. So we pick π is equal to zero,
one, two, and three to get that π‘ is equal to a quarter, five-quarters,
nine-quarters, or thirteen-quarters. Our particle will be at the point
zero, one when both the π₯-coordinate is equal to zero and the π¦-coordinate is
equal to one.

What we have done is we have found
all the values of π‘ between zero and four inclusive which give an π₯-coordinate of
zero and all the values of π‘ between zero and four inclusive which gives a
π¦-coordinate of one. So any values of π‘ which are in
both of these lists will give us an π₯-coordinate of zero and a π¦-coordinate of
one. So we just check which values of π‘
are in both of these lists. And we get that if π‘ is equal to a
quarter, five-quarters, nine-quarters, or thirteen-quarters, then the position of
the particle will be zero, one.