Trig Functions and the Unit Circle

Question

determine the missing coordinate of the point (x, -1/4) that lies on the graph of the unit circle in quadrant III.

Answer 1

For the unit circle we know that

x^2 + y^2 = 1

Here we have y = -1/4 so we can write

x^2 + (-1/4)^2 = 1

from which

x^2 + (1/16) = 1

x^2 = 1 - (1/16) = 15/16

x = +/- sqrt(15/16) = +/- (1/4) * sqrt(15)

To be in the third quadrant x should be < 0 so

x = -(1/4) * sqrt(15)