Exercise 3 Page 460

We are asked to find the exact coordinates of the point (x,y) on the unit circle. Looking at the graph, we see that (x,y) is a point of intersection of a unit circle and the terminal side of an angle θ= 225^(∘).

To find the coordinates of the point, we first need to find a relationship between x and y. Then we will substitute it into the following equation of a unit circle. This equation describes all the points that lie at distance of 1 from the origin. x^2+y^2=1 To begin, notice that the angle θ can be thought of as a sum of two angles, a straight angle and an angle of 225^(∘)-180^(∘) = 45^(∘).

Now, the terminal side of θ is the diagonal of a square since it bisects a right angle. We can draw that square by adding two straight segments connecting the point (x,y) to both of the axes.

Recall that, by definition, in a square all sides are the same length. In particular, the sides representing |x| and |y| are congruent.

As a result, |x|=|y| and the point (x,y) lies on either of the two following lines. y = x or y =- x Taking into consideration that the point (x,y) lies in the 3^(rd) Quadrant, both x and y must be negative. Therefore, their signs are the same and (x,y) lies on the first line. y=x We can substitute this relationship into the equation of a unit circle and solve for x.

We got two solutions, a positive and a negative one. Since we earlier established that x is negative, x= - 1sqrt(2) is our solution. Also y=x, so we instantly know the value of y. y=x ⇔ y = - 1/sqrt(2) As a last step, we can rationalize the denominator.

This gives us the following values of x and y. x=- sqrt(2)/2 and y =- sqrt(2)/2 The coordinates of the point (x,y) are (- 1sqrt(2),- 1sqrt(2)), which can also be written as (- sqrt(2)2,- sqrt(2)2).