College Trigonometry, 2011a

11.5 Graphs of Polar Equations 955
( √ )
get only one intersection point which can be represented by 3 2
2 , π 2
. We now investigate
other representations for the intersection points. Suppose P is an intersection point with
a representation (r, θ) which satisfies r =3sin ( θ
2)
and the same point P has a different
representation (r, θ +2πk) for some integer k which satisfies r = 3cos ( θ
2)
. Substituting
into the latter, we get r =3cos ( 1
2 [θ +2πk]) =3cos ( θ
2 + πk) . Using the sum formula for
cosine, we expand 3 cos ( θ
2 + πk) =3cos ( (
θ
2)
cos(πk) − 3sin
θ
(
2)
sin (πk) =±3cos
θ
2)
,since
sin(πk) = 0 for all integers k, andcos(πk) =±1 for all integers k. If k is an even integer,
we get the same equation r =3cos ( (
θ
2)
as before. If k is odd, we get r = −3cos
θ
2)
. This
latter expression for r leads to the equation 3 sin ( (
θ
2)
= −3cos
θ
(
2)
,ortan
θ
( 2)
= −1. Solving,
we get θ = − π √ )
2 +2πk for integers k, which gives the intersection point 3 2
2 , − π 2
. Next,
we assume P has a representation (r, θ) which satisfies r =3sin ( θ
2)
and a representation
(−r, θ +(2k +1)π) which satisfies r =3cos ( θ
2)
for some integer k. Substituting (−r) for
r and (θ +(2k +1)π) inforθ into r =3cos ( (
θ
2)
gives −r =3cos
1
2 [θ +(2k +1)π]) . Once
again, we use the sum formula for cosine to get
cos ( ( )
1
2 [θ +(2k +1)π]) = cos θ
2 + (2k+1)π
2
= cos ( ) (
θ
2 cos (2k+1)π
2
= ± sin ( )
θ
2
(
(2k+1)π
2
)
− sin ( ) ( )
θ
2 sin (2k+1)π
2
)
( )
where the last equality is true since cos =0andsin (2k+1)π
2
= ±1 for integers k.
Hence, −r =3cos ( 1
2 [θ +(2k +1)π]) can be rewritten as r = ±3sin ( θ
( )
2)
. If we choose k =0,
then sin (2k+1)π
2
=sin ( ) (
π
2 = 1, and the equation −r =3cos 1
2 [θ +(2k +1)π]) in this case
reduces to −r = −3sin ( (
θ
2)
,orr =3sin
θ
2)
which is the other equation under consideration!
What this means is that if a polar representation (r, θ) for the point P satisfies r =3sin( θ 2 ),
then the representation (−r, θ + π) forP automatically satisfies r =3cos ( θ
2)
. Hence the
equations r =3sin( θ 2 )andr = 3 cos( θ 2
) determine the same set of points in the plane.
Our work in Example 11.5.3 justifies the following.
Guidelines for Finding Points of Intersection of Graphs of Polar Equations
To find the points of intersection of the graphs of two polar equations E 1 and E 2 :
ˆ Sketch the graphs of E 1 and E 2 . Check to see if the curves intersect at the origin (pole).
ˆ Solve for pairs (r, θ) which satisfy both E 1 and E 2 .
ˆ Substitute (θ +2πk) forθ in either one of E 1 or E 2 (but not both) and solve for pairs (r, θ)
which satisfy both equations. Keep in mind that k is an integer.
ˆ Substitute (−r) forr and (θ +(2k +1)π) forθ in either one of E 1 or E 2 (but not both)
and solve for pairs (r, θ) which satisfy both equations. Keep in mind that k is an integer.