Unformatted text preview: tely many solutions to tan(θ) = 3,
and θ = 3 is only one of them!), we also went from x = 3, in which x cannot be 0, to y = x 3 in which we assume
x can be 0.
11 792 Applications of Trigonometry
algebraic maneuvers which may have altered the set of points described by the original
equation. First, we multiplied both sides by r. This means that now r = 0 is a viable
solution to the equation. In the original equation, r = 1 − cos(θ), we see that θ = 0 gives
r = 0, so the multiplication by r doesn’t introduce any new points. The squaring of
both sides of this equation is also a reason to pause. Are there points with coordinates
(r, θ) which satisfy r2 = r2 + r cos(θ) but do not satisfy r = r2 + r cos(θ)? Suppose
(r , θ ) satisﬁes r2 = r2 + r cos(θ) . Then r = ± (r )2 + r cos(θ ) . If we have that
r = (r )2 +r cos(θ ), we are done. What if r = − (r )2 + r cos(θ ) = −(r )2 −r cos(θ )?
We claim that the coordinates (−r , θ + π ), which determine the same point as (r , θ ),
satisfy r = r2 + r cos(θ). We substitute r = −r and θ = θ + π into r = r2 + r cos(θ) to
see if we get a true stat...
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