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## Algebra II (2018 edition)

### Course: Algebra II (2018 edition) > Unit 9

Lesson 3: The graphs of sine, cosine, and tangent# Intersection points of y=sin(x) and y=cos(x)

Sal draws the graphs of the sine and the cosine functions and analyzes their intersection points. Created by Sal Khan and Monterey Institute for Technology and Education.

## Want to join the conversation?

- Im a little confused as how Sal knew that the inbetween point between π and 3π/2 was 5π/4?(50 votes)
- Take the average: (π + 3π/2)/2

= (2π/2 + 3π/2)/2

= (5π/2)/2

= 5π/4(102 votes)

- Why does trigonometry use Greek letters (like alpha, beta, theta) rather then the common variable names (like x, y, n, or z)?(15 votes)
- Many parts of Mathematics use different letters. For instance, algebra uses mostly x, y, z and sometimes w. However, in number theory, mostly n and m are used. In vectors, the usage of i, j, k is very common, and when defining lines, I remember usually using r and s. When expressing prime numbers, it's almost always p and q, and with functions, f, g and h. When speaking about points, it's always A, B and C, sometimes D, E, F, G, M, N many times P, Q, R, S, T, for centers it is O, G, H, I, and some even usually add X and Y.

For parameters, it is not unusual using lambda and mu, and for derivatives it is delta. And I guess that for angles, we usually use alpha, beta, and gamma, although some use instead theta and phi.

I don't think there is much meaning for the usage of letters in distinct areas, but it is sometime easy to remember them that way, as they are usually used in similar fashion. For example, when you see point O, it is probably the center of a circle, when you see n and m, it is probably either a series or a number theory problem, if it's x, y and z it is either algebra or analytic geometry, and when you see alpha, beta and theta you can easily suppose that it s an angle problem.(78 votes)

- I'm curious as to whether there is some relationship of quadratics going on here with sine and cosine functions. When I look at the graphed functions, they resemble symmetrical parabolas that you would typically find with a quadratic function that just loop and keep going. Does anyone know if this is a coincidence or is there actually some relationship with trig functions like sine and cosine and the squares of some numbers?(12 votes)
- That's an interesting observation. Unfortunately, in nature there are a lot of things that look a little like parabolas that actually aren't, and the sine wave is one. Strangely enough, another one that isn't is if you hang a chain between two points it looks like it hangs like a parabola, but it's actually a different curve called a catenary.(41 votes)

- I don't really understand what Sal did at7:50with the root of two. He said 'we can rationalise the denominator here', could anybody explain what he does?(2 votes)
- Rationalising the denominator means removing the radical sign. The radical sign is that which is used to denote square root. We multiply both the numerator and the denominator by a number such that the denominator has no radical sign. Multiplying both the numerator and the denominator by the same number doesn't change the value of the number. For example, 2/3 = 4/6. So here, if we multiply both the numerator and the denominator by sqrt2, the denominator becomes 2 and the radical sign is removed.

At7:49, you can see that 1/sqrt2 has been converted into sqrt2/2

Just for your information, sqrt2/2 = 0.7071 and you get the same answer when you divide one by sqrt2. This kind of verifies that we have not changed the value of the expression by rationalising the denominator.(25 votes)

- is it necessary to remove a radical sign from denominator(15 votes)
- It is correct both ways, but without the radical on the bottom, it is easier to manipulate the result for other uses. (ex. a right triangle and trying to find sec, csc, and cot.)(4 votes)

- I am really puzzeled about the7:26-7:35portion of the video where Sal goes from a^2 =1/2 to

a = 1/sqrt 2....how did he do that?(7 votes)- You have 2a^2 = 1

(divide each side by 2) a^2 = 1/2

(take square root of each side) sqrt(a^2) = sqrt (1/2)

(left side) sqrt(a^2) = 1

(right side) sqrt(1/2) = sqrt(1)/sqrt(2); sqrt(1) = 1, sqrt(2) = sqrt(2) so sqrt(1)/sqrt(2) = 1/sqrt(2)

giving us..............a = 1/sqrt(2)

Then you multiply each side by sqrt(2)/sqrt(2) (which is 1)

a * sqrt(2)/sqrt(2) = a

1/sqrt(2) * sqrt(2)/sqrt(2) = (1*sqrt(2))/(sqrt(2) * sqrt(2)) = sqrt(2)/2

thus

a = sqrt(2)/2(9 votes)

- Where is 'tan' in all of this? I haven't seen it once yet in this section, why is that? I sort of get that the X axis represents 'cos' and the Y axis represents 'sin', but what about tan? Thank you.(9 votes)
- Tangent is the side located at the opposite of angle A over the adjacent side of angle A in a triangle. For example, if there is a right triangle with sides 1,1,root2 with angles 45,45,90 degrees, the tangent of 45 degrees will be 1/1 and therefore 1.(4 votes)

- I don't really understand why Sal, in these videos, uses pi and radians instead of 90, 180, 270, etc. degrees. Is there a difference between the two and is on better than the other for this topic?(3 votes)
- Remember that Trig functions have a lot to do with the unit circle and we tend to use radians there because radians use pi and you're dealing with a circle. Radians, although they don't seem so at first, are more convenient when dealing with the unit circle. Khan has made an entire playlist devoted to radians but the first few videos should sum it up pretty well: https://www.khanacademy.org/math/trigonometry/basic-trigonometry/radians_tutorial/v/introduction-to-radians(10 votes)

- At03:36isnt he marking wrong? He is marking cosine on the Y axis, but on the begining of the video he said cosine of data is the X, axis! And sine was the Y axis! Now im really confused!(5 votes)
- I understand the confusion. When he said cosine of theta is the x-axis, he was basically saying "Instead of the x-axis and integers (i.e. -2, -1, 0, 1, 2 ect...) we are going to call it the theta-axis and use radians." Also, instead of saying cos(theta) = y-axis he mixed up the measurement for the two axis. One axis has been turned into radians (theta, cis(theta)), the other has been left in integer number form. (x/y form).(7 votes)

- At9:55Sal says the triangles are congruent, but why doesn't this happen in the other two quadrants and the graph intersect 4 times? Can you not form isosceles right triangles in those quadrants? Or is there a point on the graph when the two functions are the same distance above or below the x axis?(2 votes)
- The issue is that in quadrants II and IV, one is negative while the other is positive. In order for sin(theta)=cos(theta) both the x and y values must be equal, rather than have the same absolute value. Same goes for the next question, while there are other points that are equidistant, you are looking for angles where x=y because x=cos(theta) and y=sin(theta).

If you were to draw y= sin (x) and y= cos(x) on the domain 0<=x<=2pi, you would see that they only intersect twice. Even though both pass through sqrt(2)/2 twice and -sqrt(2)/2 twice they only intersect each other at those y-values once each.(9 votes)

## Video transcript

- [Instructor] We're
asked at how many points do the graph of y equals sine of theta and y equal cosine theta intersect for theta
between zero and two pi? And it's zero is less
than or equal to theta which is less than or equal to two pi, so we're gonna include zero and two 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 to,
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 can 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 a 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 zero. So when theta is equal to zero, 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 coordinate is that? Well that's the point one comma zero. And so based on that,
what is cosine of theta when theta is equal to zero? Well cosine of theta is one and sine of theta is going to be zero. This is the x-axis at
the point of intersection with the unit circle, this is sorry, 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 two? So pi over two, we're right over here. What is that coordinate? Well that's now x is zero, y is one. So based on that, cosine of theta is zero, and what is sine of theta? Well that's going to be one. It's the y-coordinate right over here. Now let's go all the way to pi. Now let's go to all the way to pi. We're at this point in the unit circle. What is the coordinate? Well this is negative one comma zero, so what is cosine of theta? Well it's the x-coordinate
here, which is negative one. And sine of theta's going
to be the y-coordinate, which is zero. Now let's keep going. Now we're down here at three pi over two. If we go all the way around
to three pi over two, what is this coordinate? Well this is zero negative one. Cosine of theta is the x-coordinate here, so cosine of theta is going to be zero, and what is sine of theta going to be? Well it's going to negative one. And then finally we go back to two 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 equaled radio, (laughing) equals zero radians, and so what is cosine of theta? Well that's one, and sine of theta is zero. 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 zero, and let me draw, let me mark this off, so this is going to be
when y is equal to one, and this is when y is
equal to negative one. So y equals cosine of theta, I'm gonna graph, let's
see theta equals zero, cosine of theta equals one. So cosine of theta is equal to one. When theta's equal to pi over two, cosine of theta is zero. When theta's equal to pi, cosine of theta is negative one. When theta is equal to three pi over two, cosine of theta is equal to zero. That's this right over here. And then finally when theta's two pi, cosine of theta is one again, is one again. And the curve will look something, will look something like this, my best attempt to draw it. Make it a nice smooth curve. So it's going to look something, something, something like this. The look of these curves
should look somewhat familiar at this point, so it should
look something like this. 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's equal to zero, sine theta is zero. When theta is pi over
two, sine of theta is one. When theta is equal to pi, sine of theta is zero. When theta's equal to three pi over two, sine of theta is negative one, is negative one. When theta's equal to two pi, sine of theta is equal to zero. And so the graph of sine of
theta's gonna look something, something like, something like this, my
best attempt at drawing it, is going to look something like this. So just visually, we can think about the question. At how many points do the
graphs of y equals sine of theta and y equal cosine of theta intersect for this range for theta, for theta being between zero and two 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 over the, between zero and two pi. These are cyclical graphs. If we kept going, they
would keep intersecting with each other, but
just over this two pi, just over this two pi range for theta, you would get two points of intersection. Now let's think about what they are because they look, they look to be pretty close between, between right between zero and pi over two and right between pi
and three pi over two. So let's look at our unit
circle if we can figure out what those values are. Let's it looks like, it looks
like this is at pi over four. So let's verify that. So let's think about what these
values are at pi over four. So pi over four is that angle or that's the terminal side of it, so this is pi over four. Pi over four is the exact same thing as a 45 degree angle. So let's do pi over four right over here. so we have to figure
out what this point is, what the coordinates are. So let's make this, let's make this a right triangle. It's a right triangle. And so what do we know
about this right triangle? And I'm gonna draw it right over here to make it a little clearer. This is a very typical
type of right triangle so it's good to get some
familiarity with it. So let me draw my best attempt, all right. 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, has radius one, so the length of the
hypotenuse here is one. 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 let's call these,
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 one. Or two a squared is equal to one. a squared is equal to 1/2. Take the principle root of both sides, a is equal to the square root of 1/2, which is the square root
of one which is one, over the square root of two. We can rationalize the denominator here by multiplying by square root of two over square root of two, which gives us a is equal to, in the numerator square root of two, and in the denominator square root of two times square root of two is two. So this length is square
root of two over two and this length is the same thing. So this length right over here is square root of two over two and this height right over here is also square root of two over two. So based on that, what is this coordinate point? Well it's square root of
two over two to the right in the positive direction, so x is equal to square
root of two over two, and y is square root of two over two in the upwards direction,
in the vertical direction, the positive vertical direction, so it's also square root of two over two. Cosine of theta is just the, is just the x-coordinate, so it's square root of two over two. Sine of theta's just the y-coordinate. So you see immediately that they do, they are indeed equal at that point. So at this point they are both equal to square root of two, they're both equal to
square root of two over two. Now what about this point right over here which looks right in between
pi and three pi over two? So that's going to be, so this is pi, this is three pi over two, it is right, it is right over here. So it's another pi over four plus pi, so pi plus pi over four is the same thing as four pi
over four plus pi over four. So this is the angle five pi over four. So this is five pi over four. So this is equal to five pi over four. So that's what we're trying to figure out. What are the value of these functions at theta equal five pi over four? 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 one. 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, it's actually, they're
actually congruent triangles. So hypotenuse is one, 45-45-90, we then know that the length of this side is square root of two over two and the length of this side is
square root of two over two, the exact same logic we used over here. So based on that, based on that, what is the coordinate of that point? Well let's think about the x value. It's square root of two over
two in the negative direction. We have to go square root of two over two to the left of the origin, so it's negative square
root of two over two. This point right over,
this point on the a-xis is negative square root of two over two. What about the y value? Well we had to go square
root of two over two down, in the downward direction from the origin, so it's also negative
square root of two over two. So sine of, cosine of theta's
negative square root of two over two and sine of theta's also negative square root of two over two. And so we see that we do
indeed have the same value for cosine of theta and
sine of theta right there. They're both equal to, they're both at that point equal to the negative, the negative
square root of two over two.