WEBVTT
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Hello.
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It's your turn on the
function arctangent.
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OK.
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Before to start
with this exercise,
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let us remind something about
the function arctangent.
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OK.
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The function arctangent is
defined on the whole set
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of real numbers.
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And its range is equal to
the open interval minus pi
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over 2, pi over 2.
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More, this is a
bijective function.
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That is, it is both
injective and surjective.
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Good.
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Let us start with the first
point of our exercise.
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OK, we want to
compute the arctangent
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of minus the square root of 3.
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Good.
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Let us denote by
theta this value.
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Then, what is theta?
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Theta is the unique value
inside this open interval, where
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the tangent of theta is equal
to minus the square root of 3.
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OK.
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But now, we remember
that the tangent of theta
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is the sine of theta
over the cosine of theta.
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And we are looking for what?
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For a value of theta
inside this interval,
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such that this fraction is equal
to minus the square root of 3.
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But now, because this
fraction is negative,
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surely theta will be a value
between minus pi over 2 and 0.
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And precisely, which
is the correct value?
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Theta equal to minus pi over 3.
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Indeed, if you consider theta
equal to minus pi over 3,
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then what we have
that the sine of theta
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over the cosine of theta is
equal to minus the square root
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of 3/2 over 1/2.
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That is, minus the
square root of 3.
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Exactly what we wanted.
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Therefore, the arctangent of
minus the square root of 3
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is minus pi over 3.
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OK.
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And now, let us consider the
second point of our exercise.
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We want to compute the
arctangent of the sine of pi
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over 2.
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Good.
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Let us denote by theta
again this value.
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But the sine of pi
over 2 is equal to 1.
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Therefore, what is theta
is the arctangent of 1.
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OK.
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Then by the definition of
the function arctangent,
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what we have, that 1 has to be
equal to the tangent of theta.
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But the tangent of
theta is, by definition,
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the sine of theta over
the cosine of theta.
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OK.
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Therefore, theta
has to be a value
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inside this open interval,
where the sine and the cosine
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coincide.
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Which is the unique
value inside the interval
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where this happens.
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We immediately get that theta
has to be equal to pi over 4.
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Indeed, in pi over 4, both the
sine and the cosine functions
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are equal to the square
root of 2 over 2.
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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 sine
of the arctangent of 2.
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OK.
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Let us denote by theta
the arctangent of 2.
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OK.
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Therefore what we have?
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We have that the tangent
of theta is equal to 2.
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OK.
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But, what is the
tangent of theta?
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We know then that the 2
equal to the tangent of theta
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has to be equal to the sine of
theta over the cosine of theta.
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And now, attention.
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Theta is the arctangent
of something.
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Therefore, theta lives
inside this open interval.
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And for each angle inside
this open interval,
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the cosine is always
a positive number.
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OK?
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Then, what we can write here?
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OK.
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We'll leave the sine
of theta on the top.
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And write the square root
plus the square root of 1
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minus the sine squared of theta.
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Not only-- and also, you
see from this equality,
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what do we get?
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You see, this is a
positive number--
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the square root.
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This is a positive number.
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Then also, this has to
be a positive number.
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Then, not only we
have this relation,
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but also we get,
from this relation,
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that the sine of theta has
to be a positive number.
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Good.
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And now, from this
equality, what do we get?
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We take the square
of this and this.
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And then we get 4 equal to
sine squared of theta over one
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minus sine squared of theta.
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That is, multiplying on both
sides by 1 minus sine squared,
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we get 4 minus 4 times
sine squared of theta
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equal to sine squared of theta.
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That is, 5 times sine
squared of theta equal to 4.
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Which implies sine squared
of theta equal to 4/5.
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And now, because we know
that the sine of theta
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is greater than 0, we can
conclude that the sine of theta
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is equal to 2 over
the square root of 5.
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Good.
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Then, we have that
the sine of theta,
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that is the sine of the
arctangent of 2, that
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was exactly what
we had to compute,
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is equal to 2 over
the square root of 5.
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Thank you very much
for your attention.
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