Example 1:

The terminal point P(x,y) determined by t.=./4 is the same distance from (1,0) as from (0,1) along the unit circle. Since the unit circle is symmetric with respect to the line y.=.x, it follows that P lies on the line y.=.x. So P is the point of intersection of the circle x2.+.y2.=.1 and the line y.=.x. Substituting x in for y in the equation of the unit circle, we get:

x2 + x2 = 1
2x2 = 1
x2 = 1/2 x = 1/ = /2
Since P is inthe first quadrant, x = /2 and since y.=.x, we have y.=./2 also. Thus the terminal point determined by /4 is P(/2,/2).

Example 2:

Let's find the terminal point P(x,y) determined by t.=./6. Let Q(x,-y) and R(0,1) be as shown in the diagram:

The arecs PQ and PR are equal length (they both have length /3) and so they are subtended by equal chords, that is |PQ|.|PR|. Thus

(distance formula)

4y2 = x2 + y2 - 2y + 1 ________________(Squaring both sides)
4y2 = 1 - 2y + 1 _____________________(x2+y2=1)
2y2 + y - 1 = 0 ______________________(divide by 2)
(2y-1)(y+1) = 0
y = 1/2 or y = -1

Since P is in quadrant I, it follows that y.=.1/2. To find x, we again use the fact that P(x,y) is on the unit circle, that is, x2.+.y2.=.1. from this we get:
x2 + (1/2)2 = 1
x2 = 3/4
x = /2 (because x.>.0)
So we have found that the termnial point determined by t = /6 is the point P(/2,1/2).

Example 3:

We now use symmetry porperties of the circle to find the terminal point for t.=./3 using example 2.

From the diagram we see that P and Q are symmetrically placed about the line y.=.x.
Since Q.=.Q(/2,1/2), it follows that the terminal point P.=.P(1/2,/2).