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
the line y.=.x.
Substituting x in for y in the equation of the unit circle, we get:
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:
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
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
Since Q.=.Q(/2,1/2), it follows that the terminal point P.=.P(1/2,/2).
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