The point at which the terminal side of the angle [tex]\frac{7\pi}{4}[/tex] intersects the unit circle is [tex](\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2})[/tex]
To get the point [tex](x,y)[/tex] where the terminal side of the angle intersects the unit circle, note that
[tex]sin(\frac{7\pi}{4})=y[/tex]
[tex]cos(\frac{7\pi}{4})=x[/tex]
since the circle is a unit circle, the radius is 1
Computing the trigonometric ratios, calculate the reference angle for [tex]\frac{7\pi}{4}[/tex][tex](2\pi-\frac{7\pi}{4})=\frac{\pi}{4}[/tex] (The angle is in the 4th Quadrant)
[tex]y=sin(\frac{7\pi}{4})\\=-sin(\frac{\pi}{4})\\=-\frac{\sqrt{2}}{2}[/tex]
and
[tex]x=cos(\frac{7\pi}{4})\\=cos(\frac{\pi}{4})\\=\frac{\sqrt{2}}{2}[/tex]
The point of intersection is [tex](x,y)=(\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2})[/tex]
Learn more about angles here: https://brainly.com/question/10390757
