Example: Intersection of sine and cosine | Graphs of trig functions | Trigonometry | Khan Academy

We're asked at how many points did the graph of y equals sine of theta and y equal cosine theta intersect for theta between 0 and 2 pi, and it's 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, you know the possible values for theta.

To do this, I've set up a little chart for theta, cosine theta, and sine theta. We can use this to end 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.

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 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 (1, 0). Based on that, what is cosine of theta when theta is equal to zero? 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 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 (-1, 0). So what is cosine of theta? What's the x-coordinate here, which is -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, -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 -1.

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 equaled 0 radians. And so what is cosine of theta? Well, that's 1, and sine of theta is 0.

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, let me mark this off; this is going to be when y is equal to 1, and this is when y is equal to -1. So y equals cosine of theta; I'm going to graph. 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 over 2; cosine of theta is 0. When theta is equal to pi, cosine of theta is -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. The curve will look something, look something like this. My best attempt to draw it, make it a nice smooth curve. So it's going to look something, 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 -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, 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 equal sine of theta and y equal cosine of theta intersect for this range 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 over the range from 0 to 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 2 points of intersection.

Now, let's think about what they are because they look to be pretty close between right 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; we have to figure out what this point is, what the coordinates are. So let's make this a right triangle; it's a right triangle. And so what do we know about this right triangle? I'm going to 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. 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 one, so the length of the hypotenuse here is one.

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. We could say a squared plus a squared is equal to the hypotenuse squared, which is equal to one, or 2a squared is equal to 1.

So a squared is equal to one half. Take the principal root of both sides; a is equal to the square root of one half, 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 square root of 2 over 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, in the vertical direction, the positive vertical direction.

So, 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 do, and they are indeed equal at that point. So at this point, they are 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 this is 5 pi over 4.

So this is equal to 5 pi over 4; that's what we're trying to figure out. What is it? What are the values of these functions at theta equal 5 pi over 4? Well, there are multiple ways to think about it. You can 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 can 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; they're actually congruent triangles.

So, hypotenuse is 1, 45, 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 right over this top, 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're both equal to, are both at that point equal to the negative square root of 2 over 2.