Answer:
In brief, apply the pythagorean theorem to show that the distance between the point [tex](\sqrt{2}/2,\sqrt{2}/2)[/tex] and the origin is [tex]1[/tex].
Step-by-step explanation:
The pythagorean theorem can give the distance between two points on a plane if their coordinates are known.
A point is on a circle if its distance from the center of the circle is the same as the radius of the circle.
On a cartesian plane, the unit circle is a circle
Therefore, to show that the point [tex](\sqrt{2}/2,\sqrt{2}/2)[/tex] is on the unit circle, show that the distance between [tex](\sqrt{2}/2,\sqrt{2}/2)[/tex] and [tex](0,0)[/tex] equals to [tex]1[/tex].
What's the distance between [tex](\sqrt{2}/2,\sqrt{2}/2)[/tex] and [tex](0,0)[/tex]?
[tex]\displaystyle \sqrt{\left(\frac{\sqrt{2}}{2}-0}\right)^{2} + \left(\frac{\sqrt{2}}{2}-0\right)^{2}} = \sqrt{\frac{1}{2} + \frac{1}{2}}= \sqrt{1}= 1[/tex].
By the pythagorean theorem, the distance between [tex](\sqrt{2}/2,\sqrt{2}/2) [/tex] and the center of the unit circle, [tex](0,0)[/tex], is the same as the radius of the unit circle, [tex]1[/tex]. As a result, the point [tex](\sqrt{2}/2,\sqrt{2}/2)[/tex] is on the unit circle.
