Evaluating trigonometric angles in standard position

0:00- [Instructor] A circle has its center at the origin. Angle Theta in standard position intersects the circle at point -3,7. What is the value of sin of Theta? All right, pause the video. Have a go at that before we do it together. Okay, now let's just try to visualize this.

0:18So my x axis, my Y axis, if I make a circle centered at the origin right over here, that's a nice looking circle. Oh, it's a very, it's a very heavy circle, but you get the idea. And then we have an angle Theta that intersects at the point -3,7. So let's say it is -1.

0:44Whoops. I'm using the wrong tool. Oh, I'm using the circle tool still. So we have negative one, two, three, and then positive one, two, three, four, five, six, seven. So it would be maybe at this point right over here. It's hard to see. Let me see if I can, here this point right over here, is the point -3,7.

1:10And so we have the angle Theta. So if I were to draw the angle, Theta looks like this, that is Theta. And we wanna figure out sin of Theta. Now there's a couple of ways that we can approach this. We can think about, well it's a, we could think about the unit circle definition of trig functions.

1:29But what's different about this is, this isn't a unit circle. If it was a unit circle, we could just take the Y coordinate right over here and say, oh, we're done if, if this was truly a radius of one, but this radius is going to be something else. So one option is we could figure out what that is and then divide our Y coordinate or you could say our vertical, our vertical displacement by that radius.

1:53That's one way to think about it. Another way to think about it, which will give us the exact same answer, is if we construct a right triangle right over here, and this is where we can reconcile the SOHCAHTOA definition of trig functions with the unit circle definition of trig functions.

2:09You think about the angle between the radius and the x axis, this angle here, and if we say, okay, the sign of this is opposite over hypotonus, well guess what, that's going to be in this case the magnitude of the vertical displacement divided by, once again, this radius, the hypotenuse is this radius.

2:29So in any case, we need to figure out what this radius is. Now, to do that, we can just use the Pythagorean theorem because this length is seven. This length right over here is three or three to the left. The magnitude of our horizontal displacement is three. And so using the Pythagorean Theorem, we know, let's say if we call this, well let's just call this R for radius.

2:51I'll do another color that we can see better. We know that R squared is equal to 7 squared + 7 squared. So R squared = 49 + 9. R squared = 58, or that R is equal to the square root of 58. Is equal to the square root of 58. So either way you could think about it is sin, sin of our angle, of theta, is equal to opposite over the hypotonus.

3:24Or you could view it as your Y coordinate divided by the radius of this circle. And so that either way is going to be seven over the square root of 58 and we're done.