Find the intersection point of r cos(Ø - pi/4) = (2)^½ and r cosØ = 1
Must be a couple of people in the same class!
Call phi = t for convenience.
The first equation states:
rcos(t - pi/4) = rcos(t)cos(pi/4) + rsin(t)sin(pi/4) = sqrt(2)
or after some simplifying:
rcos(t) + rsin(t) = 2
Using the second equation gives:
1 + rsin(t) = 2
rsin(t) = 1
Now, consider the two equations:
rsin(t) = 1
rcos(t) = 1
Dividing the two gives:
tan(t) = 1
t = pi/4 + n*pi (where n = 0, 1, 2, 3, ...)
Now
r = 1/cos(t)
So
r = 1/cos(pi/4 + n*pi) = 1/[cos(pi/4)cos(n*pi) - sin(pi/4)*sin(n*pi)]
Now:
cos(pi/4) = sin(pi/4) = 1/sqrt(2)
cos(n*pi) = (-1)^n
sin(n*pi) = 0
Thus
r = (-1)^n*sqrt(2)
t = pi/4 + n*pi
-Dan