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In which quadrant does π lie if sin π equals one over root two and cos π equals one over root two?
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To understand this problem, weβre actually gonna use the unit circle.
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And so, the unit circle stems from the origin.
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And itβs a circle that has a radius of one.
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Okay, but how can we use this?
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Okay, letβs, first of all, consider four points on our unit circle.
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So, we have π₯, π¦; π₯, negative π¦; negative π₯, negative π¦; and negative π₯, π¦.
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Well, if we look in the top-right quadrant where we have π₯, π¦ is our coordinate on unit circle, then if we were to consider sin π, well, sin π would be equal to the opposite over the hypotenuse.
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Well, in our situation, the opposite would just be the change in π¦, so itβd be π¦.
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And because itβs a unit circle, we know the hypotenuse is actually the radius, so itβd be equal to one.
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So, you get π¦ over one.
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So therefore, itβd just be equal to π¦.
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Okay, great, so, thatβs sin π.
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Well, if we look at the cos of π, well, the cos of π would be equal to the adjacent over the hypotenuse.
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And in this scenario, the adjacent would just be π₯ cause itβd be the change in π₯.
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And then, our hypotenuse would just be one again because the hypotenuse is the radius of the circle, which is one.
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So therefore, it would just be π₯.
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So, cos of π would just be π₯.
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Okay, great, now finally, letβs look at tan π.
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Well, tan π is just equal to the opposite over the adjacent.
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Okay, so, this would give us π¦ over π₯.
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Okay, great, so weβve now got in the top-right quadrant, our answers for sin π would just be π¦, cos of π would just be π₯, and tan π would be equal to π¦ over π₯.
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And the key thing here is that they are all positive.
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And thatβs whatβs gonna be very important when we come back to this.
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And okay, Iβm gonna quickly go around the other kind of quadrants and show you what weβll get in each of those.
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So, if we next go down to the bottom-right quadrant where we have the coordinate π₯, negative π¦, then we know this time that sin π would be equal to negative π¦, because thatβd be our opposite, divided by our hypotenuse again, which is the radius, which is one.
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So, this would be equal to negative π¦.
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Well, the cos π, this time, would be equal to, again, itβd adjacent over hypotenuse.
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So, itβd be our π₯.
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Okay, itβs the same because the π₯-coordinate is still just π₯.
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And thatβd be divided by one, our hypotenuse.
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So therefore, the cos of π would just be equal to π₯.
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And then, finally, tan π would actually be equal to negative π¦ over π₯ because itβd be our opposite, which is negative π¦, over our adjacent, which would be π₯.
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So, this would just remain as negative π¦ over π₯.
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Okay, so, weβve got these three now, right.
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So, whatβs the important thing here?
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Well, itβs important to note here that actually itβs only the cos of π thatβs positive.
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The other two give us both negative answers.
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Okay, thatβs great!
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Two quadrants down.
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Letβs move on to the bottom-left quadrant.
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Well, in the bottom-left quadrant, we have the point on the unit circle negative π₯, negative π¦.
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And we can see that here, this time, what weβll have is that sin π is gonna be equal to, again, itβs gonna be the opposite over the hypotenuse.
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So, itβs negative π¦ over one, which, once again, will give us negative π¦.
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But cos of π is gonna be equal to negative π₯ over one because this time the π₯ is also negative.
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And thatβs going to give us negative π₯ as its answer.
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And finally, we take a look at tan π, well, this time, a little bit different because tan π again is going to be the opposite over the adjacent, whichβd be negative π¦ over negative π₯.
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And actually, negative π¦ over negative π₯ will give us just π¦ over π₯.
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And this is key because this is positive.
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So, we can see actually in the bottom-left quadrant, only tan π would be positive.
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Okay, great, final one, letβs go to the top-left quadrant.
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Well, here, we have the coordinates negative π₯, π¦ on our unit circle.
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So therefore, sin π is gonna be equal to π¦ over one.
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So, itβll just be equal to π¦.
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Cos π will be equal to negative π₯ over one.
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So, this time itβs gonna be equal to negative π₯.
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And tan π will just be equal to π¦ over negative π₯.
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Okay, great, and if we look at this, we can see that itβs only sin π thatβs actually going to be positive.
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Okay, great, weβve got them all completed, but why is this useful?
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Well, actually, if you have to take a look around, what weβve actually shown and proved is actually the cast diagram.
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And this is the diagram we actually use in trigonometry to show us where the trig values are going to be positive or negative.
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So, we can see top right, all of them are positive.
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Top left, only the sin π is gonna be positive.
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Bottom left, only tan π is positive.
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And bottom right, only cos of π is positive.
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Okay, great, so now that we have this, we can use it to solve our problem.
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Well, if we look at the values that weβve been given in the question, we see that sin π is equal to one over root two.
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And cos π is also equal to one over root two.
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So therefore, we know that actually both of them are positive.
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And as sin π and cos π are both positive, then π must lie in the top-right quadrant because thatβs where all of our trig ratios are positive.
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So therefore, we can say that π must lie in the first quadrant.
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And we know itβs the first quadrant because our quadrants are actually labelled from one to four anticlockwise.
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So, starting in the top right, so we have one, top left two, bottom left three, and bottom right four.
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So, yes, our π is in the first quadrant.