a special right triangle drawn in quadrant I intersects the unit circle at point (x,y) where y = √2/2. what is the value of x in point (x,y)? use the equation of the unit circle to determine the missing coordinate?
The unit circle with radius 1 is defined by the equation x^2 + y^2 = 1.
Since y = √2/2, we can substitute this value into the equation to solve for x:
x^2 + (√2/2)^2 = 1
x^2 + 2/4 = 1
x^2 + 1/2 = 1
To isolate x, we subtract 1/2 from both sides:
x^2 = 1 - 1/2
x^2 = 1/2
To find x, we take the square root of both sides:
x = √(1/2)
Simplifying the square root:
x = √(1)/√(2)
Rationalizing the denominator (√(2) * √(2) = 2):
x = √(1)/√(2) * √(2)/√(2)
x = √(2)/2
Therefore, the value of x in the point (x, y) is √(2)/2.