Cosine, sine and tangent of π/6 and π/3

In this video, we're going to figure out what the sine, cosine, and tangent of two very important angles are. Angles that you'll see a lot in your trigonometric studies, and just in general, in your regular life. So these are the angles pi over 3 radians and pi over 6 radians. Sometimes it's useful to visualize them as degrees. Pi over 3, you might remember, pi radians is 180 degrees, so you divide that by 3. This is equivalent to 60 degrees. And once again, 180 degrees, which is the same thing as pi radians divided by 6, is the same thing as 30 degrees.

Now I'm going to do it using the unit circle definition of trig functions. But to help us there, I'm going to give us a little bit of a reminder of what some of you might be familiar with as 30-60-90 triangles, which I guess we could also call pi over 6, pi over 3, pi over 2 triangles. And so let me just draw one because this is going to be really helpful in establishing these trig functions using the unit circle definition.

So let me draw a triangle here. It's hand-drawn, so it's not as neat as it could be. So this right over here is a right angle, and let's say that this one is pi over 3 radians, which is the same thing as 60 degrees. And this one over here is pi over 6 radians, which is the same thing as 30 degrees. Now let's also say that the longest side, the hypotenuse here, has length one.

Now, to help us think about what the other two sides are, what I'm going to do is flip this triangle over this side right over here and essentially construct a mirror image. So because this right over here is a mirror image, we immediately know a few things. We know that this length right over here is going to be congruent to this length over here. Let me actually finish drawing the entire triangle; it's going to look something like this.

And since it's, once again, a reflection, this length over here is going to have length 1. This is going to be pi over 6 radians, and this is going to be pi over 3 radians. So what else do we know about this larger triangle now? Well, we know it's an equilateral triangle: all the angles pi over 3 radians, pi over 3 radians, and if you add 2 pi over 6's together, you're going to get pi over 3 as well.

So it's a 60-degree, 60-degree, 60-degree triangle, and so all the sides are going to have the same length. It's going to be one, one, and one. And if these two sides are congruent to the smaller triangles, or the smaller right triangles, well then this right over here must be one half, and then this right over here must be one half as well.

Now that's going to be useful for figuring out what this length right over here is going to be because we have two right triangles. We could use either one, but if we just use this bottom right triangle here, the Pythagorean theorem tells us that one half squared, let's call this b, plus b squared — I'm just pattern matching a squared plus b squared is equal to c squared, where c is the length of the hypotenuse — is equal to 1 squared.

And so, we get that 1 over 4 plus b squared is equal to 1, or subtracting 1 over 4 from both sides, b squared is equal to 3 over 4. And then taking the principal root of both sides, we get b is equal to the square root of 3 over 2. So just like that, we have figured out what all the lengths of this 30-60-90 triangle are. So b here is equal to square root of 3 over 2.

Now I said this would be useful as we go into the unit circle definitions of sine, cosine, and tangent, and we're about to see why. So here I have two different unit circles. I'm going to use one for each of these angles. So first, let's think about pi over 3 radians. And so pi over 3 would look something like this.

So this is pi over 3 radians, and the cosine and sine can be determined by the x and y coordinates of this point where this radius intersects the actual unit circle. The coordinates here are going to be cosine of pi over 3 radians and sine of pi over 3 radians. Or another way to think about it is I can set up a 30-60-90 triangle here, so I'm going to drop a perpendicular. This would be 90 degrees, or pi over 2 radians, and then this angle over here, if this is 60, this is 90, this is going to be 30.

Or another way of thinking about it's going to be pi over 6 radians. It's going to be just like one of these triangles here. And so the x coordinate, which is going to be the same thing as the cosine of pi over 3, is going to be the length of this side right over here. Well, what's that going to be? Well, when your hypotenuse is 1, we know that the shorter side, the side opposite the pi over 6 radians, is one-half.

So just like that, we have been able to establish that cosine of pi over 3 radians is equal to one-half. This right over here is one-half. That is the x-coordinate where this radius intersects the unit circle. Now, what about the y-coordinate? What is sine of pi over 3 going to be? Well, the y coordinate is the same thing as the length of this side.

And once again, it goes back to being this triangle. If this is one, this is one-half, this is one, this is one-half, this other side is going to be square root of 3 over 2. So sine of pi over 3 is going to be square root of 3 over 2. So let me write that down: sine of pi over 3 is equal to square root of 3 over 2.

And these are good ones to know. I never say really memorize things; it's always good to know how to derive things in case you forget. But if you had to memorize some, I would highly recommend memorizing these. And then, of course, from these, we can figure out the tangent. The tangent is just going to be the sine over the cosine. So let me write it down here.

The tangent of pi over 3 is going to be the sine, which is square root of 3 over 2, over the cosine, which is one-half. Got a little scrunchy down there. And so this is just going to be square root of three over two times two, which is just going to be square root of three.

So now let's just use that same logic for pi over 6. And in fact, I encourage you to pause this video and see if you can do that on your own. All right, now let's draw a radius that forms a pi over 6 radian angle with a positive x-axis. Might look like that. So if that's going to be pi over 6 radians, you might imagine it's interesting to drop a perpendicular here and see what type of triangle we've constructed.

So this has length 1, this is pi over 6 radians, this is a right angle, so this again is going to follow the same pattern. This will be pi over 3 radians, and so the sides are exactly the exact same as this top blue triangle here. So we know that this length over here is going to be one-half. We know that this length over here is going to be square root of 3 over 2.

And that's useful because that tells us the coordinates here. The coordinates here, the x-coordinate of this point where the radius intersects the unit circle, is square root of 3 over 2, and then the y-coordinate is one-half. And that immediately tells us the cosine and the sine of pi over 6. Let's just write it down.

So this tells us that cosine of pi over 6 is equal to square root of 3 over 2, and sine of pi over 6 is equal to one-half. Notice we just actually just swapped these two things around because now the angle that we're taking the sine or cosine of is a different angle on a 30-60-90 triangle, but we're essentially utilizing the same side measures is one way to think about it.

And then what's the tangent going to be? I'll write it down here. The tangent of pi over 6 is going to be the sine over the cosine, square root of 3 over 2. And so that's going to be equal to one-half times 2 over the square root of 3, which is equal to 1 over the square root of 3.

Now some people sometimes don't like radicals in the denominator, and so you could multiply the numerator and the denominator by square root of three if you like to get something like this. You multiply the numerator and denominator by square root of three; you get square root of three over three, which is another way of writing tangent of pi over 6. But either way, we're done. It's very useful to know the cosine, sine, and tangent of both pi over 3 and pi over 6, and now you also know how to derive it.