Welcome back, everyone. So up to this point, we've been talking about special case right triangles, and we recently took a look at the 30, 60, 90 case. Well, in this video, we're going to be taking a look at the trigonometric functions associated with the 30, 60, 90 triangle. Now so far, the trigonometric functions we've learned about have been a bit tedious just because we have to find ratios for each of the situations and see how they relate to each other. But what we're hopefully going to figure out in this video is that for these special case triangles, there are patterns that show up when you solve for the trigonometric ratios and function. So without further ado, let's get right into this and see if we can find some shortcuts and patterns that show up.
So we're going to start with finding the sine of a 30, 60, 90 triangle, and we're going to focus on the 30 degree angle for these trigonometric functions on the left side. Now recall that sine is opposite over hypotenuse. So if I go to our 30 degree angle, the opposite side of this triangle is 5, and the hypotenuse or the long side of the triangle is 10. And 5 over 10 reduces to \( \frac{1}{2} \), meaning that the sine of 30 degrees is \( \frac{1}{2} \). But now let's take a look at the cosine of our 30 degree angle. For the cosine, it's going to be adjacent over hypotenuse. And the adjacent side to the 30 degree angle is \( 5 \times \sqrt{3} \), and then this is divided by the hypotenuse which is 10. And again, \( \frac{5}{10} \) reduces to \( \frac{1}{2} \), so we'll end up with \( \frac{1 \times \sqrt{3}}{2} \), and \( 1 \times \sqrt{3} \) is just \( \sqrt{3} \), so all we're gonna end up with is \( \frac{\sqrt{3}}{2} \), and that's our cosine. But now let's take a look at the tangent. For the tangent, it's going to be opposite over adjacent. Well, going to this triangle, we can see that the opposite side is 5, and then the adjacent side is \( 5 \times \sqrt{3} \). Again, the fives will cancel giving us \( \frac{1}{\sqrt{3}} \). And if I go ahead and rationalize this denominator by multiplying the top and bottom by \( \sqrt{3} \), we'll get the square roots to cancel, giving us \( \frac{\sqrt{3}}{3} \) as our tangent. So this is what we get for the sine, cosine, and tangent of our 30, 60, 90 triangle when specifically looking at the 30 degree angle.
But now let's solve for the cosecant, and we're going to use reciprocal identities to solve these next 3 trigonometric functions. Because I noticed for the cosecant, it's the same thing as \( \frac{1}{\sin(\theta)} \). And we said that the sine of theta is \( \frac{1}{2} \). So all we need to do is flip \( \frac{1}{2} \), which is just going to be 2. So the cosecant of our 30 degree angle is 2. Now let's take a look at the secant of our angle. Well, for the secant, we can see that this is \( \frac{1}{\cos(\text{angle})} \), and we already said that the cosine is \( \frac{\sqrt{3}}{2} \). So for secant, we're going to get \( \frac{2}{\sqrt{3}} \), which I will need to rationalize this denominator, so multiply the top and bottom by \( \sqrt{3} \), cancel the square roots giving us \( \frac{2 \times \sqrt{3}}{3} \), which is the result for our secant of 30 degrees.
But now let's take a look at our cotangent. For the cotangent, it's going to be \( \frac{1}{\tan(\text{angle})} \). We figured out that the tangent is \( \frac{\sqrt{3}}{3} \). So the cotangent is going to be \( \frac{3}{\sqrt{3}} \), just flipping this fraction. And again, I'll need to rationalize this denominator. When doing this, the square roots will cancel, giving us \( \frac{3 \times \sqrt{3}}{3} \), and these threes are going to cancel leaving us with just \( \sqrt{3} \). That is how you can find the trigonometric functions for the 30 degree angle. Now because this is a 30, 60, 90 triangle, we also need to find all the trigonometric functions for 60-degree angle. So let's go ahead and do that on this right side.
So for the sine function, if we're looking for 60 degrees, recall that this is opposite over hypotenuse. Now the opposite side is \( 5 \times \sqrt{3} \), and this is going to be divided by the hypotenuse, which is 10. The \( \frac{5}{10} \) will reduce to give us just a 2 in the denominator, meaning that all we're going to have is \( \frac{\sqrt{3}}{2} \) for our sine. Now let's take a look at our cosine. For the cosine of 60 degrees, recall that this is adjacent over hypotenuse, so we're going to go to the adjacent side of the 60 degree angle which is 5, and then divide this by the hypotenuse which is 10. \( \frac{5}{10} \) reduces to \( \frac{1}{2} \), meaning that the cosine of the 60 degree angle is \( \frac{1}{2} \). Now let's take a look at the tangent, which is opposite over adjacent. The side opposite to 60 degrees is \( 5 \times \sqrt{3} \), and then the adjacent side is 5. The fives will cancel giving us just \( \sqrt{3} \), meaning that \( \sqrt{3} \) is the tangent of 60 degrees. Now for these last 3 trigonometric functions, we're going to use the reciprocal identities.
Now the cosecant is the same thing as \( \frac{1}{\sin(\text{angle})} \). So if I want to find the cosecant of 60 degrees, I just need to flip the sine of 60 degrees, which would be \( \frac{2}{\sqrt{3}} \). Now I can go ahead and rationalize this denominator. That'll get the square roots to cancel on the bottom, giving us \( \frac{2 \times \sqrt{3}}{3} \). So that is going to be the cosecant of our 60 degree angle. Now what I can also do is find the secant of our 60 degree angle. And to do this, what I need to do is it's \( \frac{1}{\cos(\text{angle})} \), so I need to go ahead and flip the cosine of 60 degrees. Since the cosine of 60 degrees is \( \frac{1}{2} \), that means that the secant of 60 degrees is going to be 2. So 2 is the secant of 60 degrees. Now lastly, we're gonna take a look at the cotangent of 60 degrees, and all I need to do is go ahead and flip the tangent that we got here. So we said the tangent is \( \sqrt{3} \), so the cotangent is gonna be \( \frac{1}{\sqrt{3}} \). I can go ahead and rationalize the denominator, getting the square roots to cancel on bottom, giving \( \frac{\sqrt{3}}{3} \) for the cotangent of 60 degrees. So these are all of the trigonometric functions evaluated for the 30, 60, 90 triangle. And you may notice that we have some patterns that emerge when we figure out what all these trigonometric values are, because notice how the sine and cosine will switch places when looking at the different angles. Because the sine of 30 is \( \frac{1}{2} \), whereas the cosine of 30 is \( \frac{\sqrt{3}}{2} \). But when we go to the 60 degree angle, the sine is \( \frac{\sqrt{3}}{2} \), and the cosine is \( \frac{1}{2} \). We see the same kind of behavior in the cosecant and secant. Since these are reciprocal identities, you just switch them when looking at the different angles. Now we also see this behavior in the tangent and cotangent. Because notice when looking at the 30 degree angle, the tangent is \( \frac{\sqrt{3}}{3} \), and the cotangent is \( \sqrt{3} \). But when looking at the 60 degree angle, the tangent is just \( \sqrt{3} \), and the cotangent is \( \frac{\sqrt{3}}{3} \). So we see the same kind of switch when looking at this trigonometric function. So hopefully this helped you to recognize some of the patterns and give you some general understanding as to how you can find the trigonometric functions for a 30, 60, 90 triangle. Hope you found this video helpful. Thanks for watching.