WEBVTT

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Hi, I'm Fiona.

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Let's have a look at Tamura 2021 paper
one and question 17.

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We're asked which of the following
sketches shows the graph of sine of X ^2

c1be9cdb-5a48-40c2-9694-cad88b95e7b6-1
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+ y ^2 = 1/2 where X ^2 + y ^2 is less
than or equal to 8 Pi.

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And then these are the sketches that were
given.

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We've tried to draw them as accurate as
we could,

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but have a look at the exam print out or
the exam paper itself to get a much more

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accurate version of these sketches.

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OK,
I'm going to just move these out of the

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way for the moment so that we can focus
on the question now.

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If we have sine of some value equals half,
then our reference angle that we're going

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to start with is π / 6.

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So we also know that X ^2 + y ^2.

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This input here has to be positive
because both of its terms are squares and

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we know that it that we have this
condition X ^2 + y ^2 is less than or

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equal to 8 Pi.

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So I'm going to draw the graph of the
sine function between zero and eight Pi.

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OK,
here's the graph of sine function between

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zero and eight Pi.

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A value of 1/2 would be about here.

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And that means that the first solution
that we're going to hit is π / 6.

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The second solution that we're going to
hit is π sorry,

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five π / 6 and then we would continue
hitting solutions all along here.

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And what we notice as we have a look at
these solutions is that they come in

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pairs that are equally spaced apart.

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And now the other thing we notice is that
X ^2 + y ^2 is a familiar,

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familiar expression in a different
context.

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So we might be familiar with the fact
that X ^2 + y ^2 = r ^2 represents a

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circle centred at the origin with a
radius R.

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And so we can think of this X ^2 + y ^2
as the square of the radius of a number

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of circles.

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OK,
and we've got these solutions pairs here.

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But in order to find the actual value of
the radius of these circles or the

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radiuses of these circles,
we need to apply the positive square root.

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Let's have a look at the graph of y =
sqrt X.

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OK, here's the graph of y = sqrt X.

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Now,
if I were to take these pairs of

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solutions that are equally spaced apart,
let's see what happens when I input them

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into the square root function.

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So when I input these pairs of equally
spaced apart solutions into the square

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root,
the graph of square root of XI can see

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that the outputs here remain in pairs,
but the gap between them.

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As their values increase,
the gap between them reduces.

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So we can think of the fact that this Y
value here is the radius of a circle.

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So we can imagine what that would look
like.

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This one is also the radius of a circle.

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Then we have another pair here,
but this pair,

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the gap between them is closer together.

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And then we have another pair here,
and the gap between them is closer

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together.

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Forgive my inaccurate drawing,
but now let's go back to looking at the

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sketches that we are given to choose from,
and we can see that every one of them

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represents a number of concentric circles,
and we know that our solutions are

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related to the radiuses of these circles.

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We also know that the radius of these
circles come in pairs,

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and so we might be looking at AB and E as
possible solutions.

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But as the value of these radiuses
increase,

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the gap between the pairs decreases,
and A is the sketch that represents this.

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So A is the solution to this question.