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# Les points d'intersection des courbes des fonctions sinus et cosinus

Transcription de la vidéo

We're asked at how many
points do the graph of y equals sine of theta and
y equal cosine of theta intersect for theta
between 0 and 2 pi. And 0 is less than or equal
to theta which is less than or equal to 2 pi. So we're going to
include 0 and 2 pi in the possible
values for theta. So to do this I've set up
a little chart for theta, cosine theta, and sine theta. And we can use this
and the unit circle to hopefully quickly
graph what the graphs of y equals sine theta and y
equals cosine theta are. And then we could think about
how many times they intersect and maybe where they
actually intersect. So let's get started. So first of all, just to be
clear, this is the unit circle. This is the x-axis. This is the y-axis. Over here we're going to
graph these two graphs. So this is going
to be the y-axis. And it's going to be a
function of theta, not x, on the horizontal axis. So first let's think
about what happens when theta is equal to 0. So when theta is equal to
0, you're at this point right over here. Let me do it in a
different color. You're at this point right
over here on the unit circle. And what to coordinate is that? Well that's the point 1 comma 0. And so based on that,
what is cosine of theta when theta is equal to 0? Well cosine of theta is 1, and
sine of theta is going to be 0. This is the x-coordinate
at the point of intersection with
the unit circle. This is the y-coordinate. Let's keep going. What about pi over 2? So pi over 2. We are right over here. What is that coordinate? Well that's now x is 0, y is 1. So based on that,
cosine of theta is 0. And what is sine of theta? Well that's going to be 1. It's the y-coordinate
right over. Now let's go all the way to pi. We're at this point
in the unit circle. What is the coordinate? Well this is negative 1, 0. So what is cosine of theta? What's the x-coordinate here? Which is negative 1. And sine of theta is going to
be the y-coordinate, which is 0. Now let's keep going. Now we're down here
at 3 pi over 2. If we go all the way
around to 3 pi over 2, what is this coordinate? Well this is 0, negative 1. Cosine of theta is
the x-coordinate here. So cosine of theta
is going to be 0. And what is sine of
theta going to be? Well it's going
to be negative 1. And then finally
we go back to 2 pi, which is making a full
revolution around the circle. We went all the way around
and we're back to this point right over here. So the coordinate is
the exact same thing as when the angle
equals 0 radians. And so what is cosine of theta? Well that's 1. And sine of theta is 0. And from this we can make
a rough sketch of the graph and think about where
they might intersect. So first let's do
cosine of theta. When theta is 0-- and
let me mark this off. So this is going to be
when y is equal to 1. And this is when y is
equal to negative 1. So y equals cosine of theta. Let's see... theta equals 0. Cosine of theta equals 1. So cosine of theta
is equal to 1. When theta is equal to pi
2, cosine of theta is 0. When theta is equal to pi,
cosine of theta is negative 1. When theta is equal to 3
pi over 2, cosine of theta is equal to 0. That's this right over here. And then finally when theta is 2
pi, cosine of theta is 1 again. And the curve will look
something like this. My best attempt to draw it. Make it a nice smooth curve. So it's going to look
something like this. The look of these curves
should look somewhat familiar at this point. So this is the graph of y
is equal to cosine of theta. Now let's do the same
thing for sine theta. When theta is equal
to 0, sine theta is 0. When theta is pi over
2, sine of theta is 1. When theta is equal to
pi, sine of theta is 0. When theta is equal to 3 pi over
2, sine of theta is negative 1. When theta is equal to 2 pi,
sine of theta is equal to 0. And so the graph
of sine of theta is going to look
something like this. My best attempt at drawing it. So just visually, we can
think about the question. At how many points do the graphs
of y equals sine of theta and y equals cosine of theta intersect
for this range for theta? For theta being between 0 and 2
pi, including those two points. Well, you just
look at this graph. You see there's two
points of intersection. This point right over here and
this point right over here. Just between 0 and 2 pi. These are cyclical graphs. If we kept going,
they would keep intersecting with each other. But just over this 2
pi range for theta, you get two points
of intersection. Now let's think
about what they are, because they look to be pretty
close between 0 and pi over 2. And right between
pi and 3 pi over 2. So let's look at our unit
circle if we can figure out what those values are. It looks like this
is at pi over 4. So let's verify that. So let's think about what
these values are at pi over 4. So pi over 4 is that angle, or
that's the terminal side of it. So this is pi over 4. Pi over 4 is the exact same
thing as a 45 degree angle. So let's do pi over
4 right over here. So we have to figure out
what this point is what. What the coordinates are. So let's make this
a right triangle. And so what do we know
about this right triangle? And I'm going to draw
it right over here, to make it a little clear. This is a typical type
of right triangle. So it's good to get some
familiarity with it. So let me draw my best attempt. Alright. So we know it's
a right triangle. We know that this is 45 degrees. What is the length
of the hypotenuse? Well this is a unit circle. It has radius 1. So the length of the
hypotenuse here is 1. And what do we know about
this angle right over here? Well, we know that it
too must be 45 degrees, because all of these angles
have to add up to 180. And since these two
angles are the same, we know that these two sides
are going to be the same. And then we could use
the Pythagorean Theorem to think about the
length of those sides. So using the
Pythagorean Theorem, knowing that these
two sides are equal, what do we get for the
length of those sides? Well, if this has length a, well
then this also has length a. And we can use the
Pythagorean Theorem. And we could say a
squared plus a squared is equal to the hypotenuse squared. Is equal to 1. Or 2a squared is equal to
1a squared, is equal to 1/2. Take the principal
root of both sides. a is equal to the
square root of 1/2 which is the square
root of 1, which is 1, over the square root of 2. We can rationalize
the denominator here by multiplying by
square root of 2 over square root
of 2, which gives us a is equal to-- in the
numerator-- square root of 2. And in the denominator,
square root of 2 times square
root of 2 is 2. So this length is
the square root of 2. And this length
is the same thing. So this length right over here
is square root of 2 over 2. And this height right
over here is also square root of 2 over 2. So based on that, what
is this coordinate point? Well, it's square
root of 2 over 2 to the right in the
positive direction. So x is equal to square
root of 2 over 2. And y is square root of 2 over
2 in the upwards direction, the vertical direction, the
positive vertical direction. So it's also square
root of 2 over 2. Cosine of theta is
just the x-coordinate. So it's square root of 2 over 2. Sine of theta is just
the y-coordinate. So you see immediately that they
are indeed equal at that point. So at this point they're
both equal to square root of 2 over 2. Now what about this
point right over here, which looks right in
between pi and 3 pi over 2. So that's going to be-- so this
is pi, this is 3 pi over 2. It is right over here. So it's another
pi over 4 plus pi. So pi plus pi over 4 is the
same thing as 4 pi over 4 plus pi over 4. So this is the
angle 5 pi over 4. So that's what we're
trying to figure out. What are the value of
these functions at theta equals 5 pi over 4? Well, there's multiple
ways to think about it. You could even use a
little bit of geometry to say, well if this
is a 45 degree angle, then this right over here
is also a 45 degree angle. You could say that the reference
angle in terms of degrees is 45 degrees. And we could do a
very similar thing. We can draw a right triangle. We know the hypotenuse is 1. We know that if this is a right
angle, this is 45 degrees. If that's 45 degrees, then
this is also 45 degrees. And we have a triangle
that's very similar. They're actually
congruent triangles. So hypotenuse is 1, 45, 90. We then know that the
length of this side is square root of 2 over 2. And the length of this side
is square root of 2 over 2. The exact same logic
we used over here. So based on that, what is
the coordinate of that point? Well, let's think
about the x value. It's square root of 2 over
2 in the negative direction. We have to go square
root of 2 over 2 to the left of the origin. So it's negative square
root of 2 over 2. This point on the x-axis
is negative square root of 2 over 2. What about the y value? Well, we have to go square
root of 2 over 2 down, in the downward direction,
from the origin. So it's also negative
square root of 2 over 2. So cosine of theta is negative
square root of 2 over 2. And sine of theta is
also negative square root of 2 over 2. And so we see that we do indeed
have the same value for cosine of theta and sine of
theta right there. They are both equal to
the negative square root of 2 over 2.