Video: Determining in Which Quadrant an Angle Lies given Trigonometric Functions

In which quadrant does πœƒ lie if sin πœƒ = 1/√2 and cos πœƒ = 1/√2?

06:08

Video Transcript

In which quadrant does πœƒ lie if sin πœƒ equals one over root two and cos πœƒ equals one over root two?

To understand this problem, we’re actually gonna use the unit circle. And so, the unit circle stems from the origin. And it’s a circle that has a radius of one. Okay, but how can we use this? Okay, let’s, first of all, consider four points on our unit circle. So, we have π‘₯, 𝑦; π‘₯, negative 𝑦; negative π‘₯, negative 𝑦; and negative π‘₯, 𝑦.

Well, if we look in the top-right quadrant where we have π‘₯, 𝑦 is our coordinate on unit circle, then if we were to consider sin πœƒ, well, sin πœƒ would be equal to the opposite over the hypotenuse. Well, in our situation, the opposite would just be the change in 𝑦, so it’d be 𝑦. And because it’s a unit circle, we know the hypotenuse is actually the radius, so it’d be equal to one. So, you get 𝑦 over one. So therefore, it’d just be equal to 𝑦. Okay, great, so, that’s sin πœƒ.

Well, if we look at the cos of πœƒ, well, the cos of πœƒ would be equal to the adjacent over the hypotenuse. And in this scenario, the adjacent would just be π‘₯ cause it’d be the change in π‘₯. And then, our hypotenuse would just be one again because the hypotenuse is the radius of the circle, which is one. So therefore, it would just be π‘₯. So, cos of πœƒ would just be π‘₯.

Okay, great, now finally, let’s look at tan πœƒ. Well, tan πœƒ is just equal to the opposite over the adjacent. Okay, so, this would give us 𝑦 over π‘₯. Okay, great, so we’ve now got in the top-right quadrant, our answers for sin πœƒ would just be 𝑦, cos of πœƒ would just be π‘₯, and tan πœƒ would be equal to 𝑦 over π‘₯. And the key thing here is that they are all positive. And that’s what’s gonna be very important when we come back to this.

And okay, I’m gonna quickly go around the other kind of quadrants and show you what we’ll get in each of those. So, if we next go down to the bottom-right quadrant where we have the coordinate π‘₯, negative 𝑦, then we know this time that sin πœƒ would be equal to negative 𝑦, because that’d be our opposite, divided by our hypotenuse again, which is the radius, which is one. So, this would be equal to negative 𝑦.

Well, the cos πœƒ, this time, would be equal to, again, it’d adjacent over hypotenuse. So, it’d be our π‘₯. Okay, it’s the same because the π‘₯-coordinate is still just π‘₯. And that’d be divided by one, our hypotenuse. So therefore, the cos of πœƒ would just be equal to π‘₯. And then, finally, tan πœƒ would actually be equal to negative 𝑦 over π‘₯ because it’d be our opposite, which is negative 𝑦, over our adjacent, which would be π‘₯. So, this would just remain as negative 𝑦 over π‘₯.

Okay, so, we’ve got these three now, right. So, what’s the important thing here? Well, it’s important to note here that actually it’s only the cos of πœƒ that’s positive. The other two give us both negative answers. Okay, that’s great! Two quadrants down. Let’s move on to the bottom-left quadrant.

Well, in the bottom-left quadrant, we have the point on the unit circle negative π‘₯, negative 𝑦. And we can see that here, this time, what we’ll have is that sin πœƒ is gonna be equal to, again, it’s gonna be the opposite over the hypotenuse. So, it’s negative 𝑦 over one, which, once again, will give us negative 𝑦. But cos of πœƒ is gonna be equal to negative π‘₯ over one because this time the π‘₯ is also negative. And that’s going to give us negative π‘₯ as its answer.

And finally, we take a look at tan πœƒ, well, this time, a little bit different because tan πœƒ again is going to be the opposite over the adjacent, which’d be negative 𝑦 over negative π‘₯. And actually, negative 𝑦 over negative π‘₯ will give us just 𝑦 over π‘₯. And this is key because this is positive. So, we can see actually in the bottom-left quadrant, only tan πœƒ would be positive. Okay, great, final one, let’s go to the top-left quadrant.

Well, here, we have the coordinates negative π‘₯, 𝑦 on our unit circle. So therefore, sin πœƒ is gonna be equal to 𝑦 over one. So, it’ll just be equal to 𝑦. Cos πœƒ will be equal to negative π‘₯ over one. So, this time it’s gonna be equal to negative π‘₯. And tan πœƒ will just be equal to 𝑦 over negative π‘₯. Okay, great, and if we look at this, we can see that it’s only sin πœƒ that’s actually going to be positive. Okay, great, we’ve got them all completed, but why is this useful?

Well, actually, if you have to take a look around, what we’ve actually shown and proved is actually the cast diagram. And this is the diagram we actually use in trigonometry to show us where the trig values are going to be positive or negative. So, we can see top right, all of them are positive. Top left, only the sin πœƒ is gonna be positive. Bottom left, only tan πœƒ is positive. And bottom right, only cos of πœƒ is positive.

Okay, great, so now that we have this, we can use it to solve our problem. Well, if we look at the values that we’ve been given in the question, we see that sin πœƒ is equal to one over root two. And cos πœƒ is also equal to one over root two. So therefore, we know that actually both of them are positive. And as sin πœƒ and cos πœƒ are both positive, then πœƒ must lie in the top-right quadrant because that’s where all of our trig ratios are positive.

So therefore, we can say that πœƒ must lie in the first quadrant. And we know it’s the first quadrant because our quadrants are actually labelled from one to four anticlockwise. So, starting in the top right, so we have one, top left two, bottom left three, and bottom right four. So, yes, our πœƒ is in the first quadrant.

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