Test whether the point is on the unit circle by plugging it into the equation, x 2 + y 2 = 1 x^2+y^2=1 x 2 + y 2 = 1 . ( − 2 2 , − 2 2 ) \left(\frac{-\sqrt2}{2},\frac{-\sqrt2}{2}\right) ( 2 − 2 <path d="M95,702 c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14 c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54 c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10 s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429 c69,-144,104.5,-217.7,106.5,-221 l0 -0 c5.3,-9.3,12,-14,20,-14 H400000v40H845.2724 s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7 c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47z M834 80h400000v40h-400000z"> , 2 − 2 <path d="M95,702 c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14 c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54 c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10 s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429 c69,-144,104.5,-217.7,106.5,-221 l0 -0 c5.3,-9.3,12,-14,20,-14 H400000v40H845.2724 s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7 c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47z M834 80h400000v40h-400000z"> )

In this problem, we're told to test whether the point is on the unit circle by plugging it into the equation, and our unit circle equation is right here, x squared plus y squared equals 1. So let's go ahead and do that. Now looking at this equation, x squared plus y squared equals 1, and the point that I'm given, I have my x value and my y value, we can just go ahead and plug those right in. Now we're going to check if it makes this equation true because if it does, we know that this point is going to be located somewhere on our unit circle. So we're testing, does x squared plus y squared equal 1 by plugging in these values? So we have negative root 2 over 2, and we want to square that, and we're adding that together with again root negative root 2 over 2 because they happen to have the same values here. Squaring that, and we want to know is this equal to 1. So let's go ahead and do some algebra here. Now squaring this fraction I need to make sure to distribute that square to both the numerator and the denominator. So squaring the numerator that negative will go away because negative one squared is a positive. So this will give me 2 because I'm really just canceling out that square root. And then in my denominator, I have a value of 4 because I'm squaring that value of 2. Now because these x and y values are the exact same, I know that my y squared will be the same thing. So I have 2 4ths plus 2 4ths, and I'm still testing, is this equal to 1? Now looking at these, adding them together, they already have the same denominator. So this gives me 4 over 4, just adding those numerators together. And we're checking, is this equal to 1? Now my knowledge of fraction kicks in here. I know that 4 over 4 is indeed 1. So this is really just saying that one is equal to 1, which is definitely a true statement. So we know that, yes, this is on the unit circle located somewhere on that. So thanks for watching, and let me know if you have questions.

3. Unit Circle

Defining the Unit Circle

Trigonometry

3. Unit Circle

Defining the Unit Circle

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Multiple Choice

Test whether the point is on the unit circle by plugging it into the equation, $x^2+y^2=1$ .
$\left(\frac{-\sqrt2}{2},\frac{-\sqrt2}{2}\right)$

On Unit Circle

NOT on Unit Circle

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6:11m

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Struggling with Trigonometry?

Test whether the point is on the unit circle by plugging it into the equation, ﻿x2+y2=1x^2+y^2=1x2+y2=1﻿.﻿(−22,−22)\left(\frac{-\sqrt2}{2},\frac{-\sqrt2}{2}\right)(2−2​​,2−2​​)﻿

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Test whether the point is on the unit circle by plugging it into the equation, $x^2+y^2=1$ .
$\left(\frac{-\sqrt2}{2},\frac{-\sqrt2}{2}\right)$