 Hello. In this short screencast for the trigonometry course, we will do one problem that uses the Pythagorean identity for the cosine and sine functions. As a bit of a clue, here's the information we will need about the Pythagorean identity. Remember that we're working on the unit circle. The equation for the unit circle is x squared plus y squared equals 1. And from our definitions, we know that x is equal to the cosine of t and y is equal to the sine of t. The value for t is the length of an arc as we wrap it around a unit circle in our usual conventions. The big, the important thing is the terminal point for that arc. So say, for example, the terminal arc is at this point right here. That means that the coordinates of that point, if we label them as x and y, then we have our definitions x equals cosine of t and y equals sine of t. This terminal point is the important thing. It doesn't matter if we kind of move around the circle like that and get to that terminal point, or if we move around the circle in a different direction, such as this, and even maybe go around more than once, but come back and have that as our terminal point. So again, it's the terminal point that's the important thing in this whole thing. And so with our trigonometric notation and so forth, we eventually end up with this as the Pythagorean identity that we use. So, here's our problem. If we know the value of the cosine of t, in this case, five-sevenths, what we want to do is determine the possible values for sine of t. I put the unit circle here because that's useful and at least visualizing what the solution to the problem would look like. Remember that cosine of t is the x-coordinate, and we're saying that that's five-sevenths. So approximately right there. And what we're asking for then are the possible y-coordinates or the possible values of sine of t. If we kind of draw in a vertical line here, we'll see that this intersects the circle in two points. And these then would be, as we move over to the y-axis, the possible values for the y-coordinate or for sine of t. So we should get two solutions for sine of t out of this. So, using our identity, our Pythagorean identity, which I've written here, we just substitute for cosine of t. And we get five-sevenths quantity squared plus sine squared of t equal to one. And now what basically our first step is to, in effect, solve for sine squared of t. So what we're going to do, of course, is square this. We get 25 over 49 plus sine squared of t equals one. Or sine squared of t equals one, which I'm going to write as 49 divided by 49 to make it easier to deal with the fractions minus 25 over 49. Or we get sine squared of t equals 24 over 49. And as the fact that we got sine squared of t is what's going to give us two solutions for sine of t. So our last steps will say sine of t will be plus or minus the square root of 24 over 49. And it's that plus or minus the square root. There are two square roots involved there. Now one last step we take most often is the fact that if we use properties of square roots, we can see that the square root of 49 is seven. So our final answer we will usually write as sine of t will be plus or minus the square root of 24 over seven. Now that result for sine of t is considered the exact value for sine of t. It involves no approximations whatsoever. The square root of 24 is a perfectly legitimate real number. So what we have is our values, our two possible values for sine of t, which we can see on the graph at these two points here. One last thing we could do, although it would not be required, is to take out our calculator and get a decimal approximation for this. If we do that, we would write this is approximately equal to 0.69985. And there you have a worked out example of a problem that uses the Pythagorean identity. So long.