If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

Main content

# Intersection points of y=sin(x) and y=cos(x)

## 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.