Example 1:

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



Example 2:

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

picture of circle

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

    SQRT[(x-x)^2+(y+y)^2] = SQRT[(x-0)^2 + (y-1)^2] (distance formula)

    SQRT(4y^2) = SQRT(x^2 + y^2 - 2y + 1)
    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 = SQRT(3)/2 (because x.>.0)
So we have found that the termnial point determined by t = pi/6 is the point P(SQRT(3)/2,1/2).



Example 3:

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

Picture of circle

From the diagram we see that P and Q are symmetrically placed about the line y.=.x.
Since Q.=.Q(SQRT(3)/2,1/2), it follows that the terminal point P.=.P(1/2,SQRT(3)/2).

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