(c) Once again, we need to manipulate
r
= 1

cos(
θ
) a bit before using the conversion
formulas given in Theorem
2.7
. We could square both sides of this equation like we did
in part
2a
above to obtain an
r
2
on the left hand side, but that does nothing helpful for
the right hand side. Instead, we multiply both sides by
r
to obtain
r
2
=
r

r
cos(
θ
).
We now have an
r
2
and an
r
cos(
θ
) in the equation, which we can easily handle, but
we also have another
r
to deal with.
Rewriting the equation as
r
=
r
2
+
r
cos(
θ
)
and squaring both sides yields
r
2
=
(
r
2
+
r
cos(
θ
)
)
2
.
Substituting
r
2
=
x
2
+
y
2
and
r
cos(
θ
) =
x
gives
x
2
+
y
2
=
(
x
2
+
y
2
+
x
)
2
.
Once again, we have performed some
algebraic maneuvers which may have altered the set of points described by the original
8
Here, ‘equivalent’ means they represent the same point in the plane.
As ordered pairs
, (3
,
0) and (

3
,
τ
2
) are
different, but when interpreted as polar coordinates, they correspond to the same point in the plane. Mathematically
speaking, relations are sets of ordered pairs, so the equations
r
2
= 9 and
r
=

3 represent different relations since
they correspond to different sets of ordered pairs. Since polar coordinates were defined geometrically to describe the
location of points in the plane, however, we concern ourselves only with ensuring that the sets of
points
in the plane
generated by two equations are the same. This was not an issue, by the way, when we first defined relations as sets of
points in the plane. Back then, a point in the plane was identified with a unique ordered pair given by its Cartesian
coordinates.
9
In addition to taking the tangent of both sides of an equation (There are infinitely many solutions to tan(
θ
) =
√
3,
and
θ
=
2
τ
3
is only one of them!), we also went from
y
x
=
√
3, in which
x
cannot be 0, to
y
=
x
√
3 in which we assume
x
can
be 0.
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Applications of Trigonometry
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
r
2
=
(
r
2
+
r
cos(
θ
)
)
2
but do not satisfy
r
=
r
2
+
r
cos(
θ
)? Suppose
(
r
0
, θ
0
) satisfies
r
2
=
(
r
2
+
r
cos(
θ
)
)
2
. Then
r
0
=
±
(
(
r
0
)
2
+
r
0
cos(
θ
0
)
)
. If we have that
r
0
= (
r
0
)
2
+
r
0
cos(
θ
0
), we are done. What if
r
0
=

(
(
r
0
)
2
+
r
0
cos(
θ
0
)
)
=

(
r
0
)
2

r
0
cos(
θ
0
)?
We claim that the coordinates (

r
0
, θ
0
+
τ
2
), which determine the same point as (
r
0
, θ
0
),
satisfy
r
=
r
2
+
r
cos(
θ
). We substitute
r
=

r
0
and
θ
=
θ
0
+
τ
2
into
r
=
r
2
+
r
cos(
θ
) to
see if we get a true statement.

r
0
?
=
(

r
0
)
2
+
(

r
0
cos(
θ
0
+
τ
2
)
)

(

(
r
0
)
2

r
0
cos(
θ
0
)
)
?
=
(
r
0
)
2

r
0
cos(
θ
0
+
τ
2
)
Since
r
0
=

(
r
0
)
2

r
0
cos(
θ
0
)
(
r
0
)
2
+
r
0
cos(
θ
0
)
?
=
(
r
0
)
2

r
0
(

cos(
θ
0
))
Since cos(
θ
0
+
τ
2
) =

cos(
θ
0
)
(
r
0
)
2
+
r
0
cos(
θ
0
)
X
=
(
r
0
)
2
+
r
0
cos(
θ
0
)
Since both sides worked out to be equal, (

r
0
, θ
0
+
τ
2
) satisfies
r
=
r
2
+
r
cos(
θ
) which
means that any point (
r, θ
) which satisfies
r
2
=
(
r
2
+
r
cos(
θ
)
)
2
has a representation
which satisfies
r
=
r
2
+
r
cos(
θ
), and we are done.