WEBVTT

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Welcome to "It's your
turn on More inverse

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trigonometric functions".

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OK.

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This exercise is dedicated
to the functions arcsin

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and arccosine.

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Let us start reminding
where they are defined

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and which is their range.

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OK.

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First, the function arcsine is
defined on the closed interval

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minus 1, 1, and its range
is the closed interval minus

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pi over 2, pi over 2.

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And the arccosine function
is defined, again,

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in the closed
interval minus 1, 1,

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but its range is the
closed interval 0, pi.

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OK?

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Good.

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Both of these functions
are bijective.

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That is, they are both injective
and surjective functions.

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OK?

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Good.

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With this in mind,
let us start to solve

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the points of our exercise.

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Let us start with the point 1.

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We have to compute
the arcsine of 1.

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OK.

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Let us denote by
theta this value.

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What means that theta
is the arcsine of 1?

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This means that the sine
of theta is equal to 1

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and theta belongs
to this interval.

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OK?

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Then which is the unique angle
inside this interval where

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the sine is equal to 1?

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Clearly it's theta
equal to pi over 2.

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Good.

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And now let us move to the
second point of this exercise.

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We have to compute the arcsine
of the sine of 3 pi over 4.

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Good.

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Again, let us denote
by theta the arcsine

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of the sine of 3 pi over 4.

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What means that the theta
is equal to this value?

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This means, again,
that the sine of theta

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is equal to the
sine of 3 pi over 4,

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but also that theta is to
belong to this interval.

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You see?

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For this reason, we cannot
declare that theta is equal

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to 3 pi over 4, because 3 pi
over 4 is not in this interval.

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OK.

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But which is the
sine in 3 pi over 4?

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3 pi over 4 is pi
minus pi over 4.

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Then this sine is equal
to the sine of pi over 4

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and the pi over 4
belongs to this interval.

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Then we immediately
get that theta,

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exactly what we were looking
for, is equal to pi over 4.

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Good.

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And now let us consider the
third point of our exercise.

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We have to compute the cosine
of the arcsine of 3 over 5.

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OK.

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Let us denote by theta
the arcsine of 3 over 5.

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OK.

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What means?

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Means that the sine of theta
is equal to 3 over 5 not

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only, and that theta
belongs to this interval.

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And now look, the cosine of
all the angles in this interval

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are greater or equal than 0.

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OK.

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Then usually the
cosine of an angle

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is plus or minus
the square root of 1

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minus sine squared
of the same angle.

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But because theta belongs
to this interval, I repeat,

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the cosine is positive.

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And then what do we get?

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That the cosine of theta is
equal to plus the square root

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of 1 minus sine
squared of theta.

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That is the square root
of 1 minus 9 over 25,

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which is equal to the
square root of 16 over 25.

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That is 4 over 5.

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Then the cosine
of theta, which is

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the cosine of the
arcsine of 3 over 5,

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exactly what we wanted to
compute, is equal to 4 over 5.

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And now let us
finish our exercise

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with this fourth point.

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What do we have?

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OK, we want to compute the
tangent of the arccosine

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of minus 2 over 3.

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Good.

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Let us denote by theta now the
arccosine of minus 2 over 3.

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OK.

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What means?

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Means that the cosine of theta
is equal to minus 2 over 3--

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not only-- and that theta
belong to this interval, 0 pi.

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OK.

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Now observe that the sine of all
the angles inside this interval

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is always greater
or equal than 0.

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Then what do we get?

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We get that the sine of
theta is a number greater

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or equal than 0.

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And then it's equal to
plus the square root

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of 1 minus the square
of the cosine of theta.

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That is the square root
of 1 minus 4 over 9,

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which is equal to the
square root of 5 over 3.

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OK.

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Now we have the cosine of theta.

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We have the sine of theta.

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We want to compute
the tangent of this,

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which is the tangent of theta.

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OK.

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We are very close to
the end of this point.

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Indeed, the tangent of theta,
which is, by definition,

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the sine of theta over
the cosine of theta,

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is equal to the square root of
5 over 3 over minus 2 over 3.

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Therefore it is equal to minus
the square root of 5 over 2.

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OK.

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[IN ITALIAN] Goodbye to everyone

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