At point [tex](\frac{\sqrt{2}}{2} ,- \frac{\sqrt{2}}{2} )[/tex] the terminal side of the angle [tex]\frac{7\pi}{4}[/tex] in standard position intersect the unit circle.
A unit circle is divided into four quadrants, each of which has an angle of 90°, 180°, 270°, and 360° (in degrees) or π/2, π. 3π/2, and 2π (in radians) respectively.
A unit circle angle is always measured from the positive x-axis, with the vertex at the origin. Its initial side is on the x-axis, while the terminal side is formed by the ray that begins at the origin and coincides with the point on the unit circle. When an angle is measured anticlockwise from the x-axis, it has a positive value.
Quadrant 1 – (0 – 90°) : X is positive, Y is positive
Quadrant 2 – (90° – 180°) : X is negative, Y is positive
Quadrant 3 – (180° – 270°) : X is negative, Y is negative
Quadrant 4 – (270° – 360°) : X is positive, Y is negative
Given,
The radius of circle = 1 unit
The terminal side of angle = \frac{7\pi}{4}
[tex]\frac{7\pi}{4}=\frac{7*180}{4}=315\textdegree[/tex]
here, the angle 315° lies in the fourth quadrant, and we know that in 4th quadrant the x is positive and y is negative.
So, from the given options, the point [tex](\frac{\sqrt{2}}{2} ,- \frac{\sqrt{2}}{2} )[/tex] lies in the 4th quadrant.
To learn more about quadrants refer here
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Your question is incomplete, here is the complete question.
At what point does the terminal side of the angle 7π/4 in standard position intersect the unit circle?
a) (-√2 / 2, -√2 / 2)
b) (-√2 / 2, √2 / 2)
c) (√2 / 2, -√2 / 2)
d) (√2 / 2, √2 / 2)