Points of Intersection of Polar Equations

May 2016
368
5
NYC
Find the points of intersection of the graphs of the polar equation.

I need a good, simple explanation.

r = 1 + cos θ

r = 1 - sin θ

I know to set both equations to equal each other. This will yield the meeting place for both equations on the graph.

1 + cos θ = 1 - sin θ

1 - 1 = - sin θ - cos θ

0 = - sin θ - cos θ

sin θ + cos θ = 0

I found this trig equation online with instructions to divide everything by cos θ. I have no idea why we must divide everything by cos θ and not sin θ. What is the reason for this division in terms of cos θ?

After dividing across by cos θ, I ended up with the following trigonometric equation:

tan θ + 1 = 0

tan θ = -1

I am stuck here.
 

HallsofIvy

MHF Helper
Apr 2005
20,249
7,909
Find the points of intersection of the graphs of the polar equation.

I need a good, simple explanation.

r = 1 + cos θ

r = 1 - sin θ

I know to set both equations to equal each other. This will yield the meeting place for both equations on the graph.

1 + cos θ = 1 - sin θ

1 - 1 = - sin θ - cos θ

0 = - sin θ - cos θ

sin θ + cos θ = 0

I found this trig equation online with instructions to divide everything by cos θ. I have no idea why we must divide everything by cos θ and not sin θ. What is the reason for this division in terms of cos θ
Dividing by cos θ is one way to solve this problem. You could also divide everything by sin θ, getting 1+ cot θ= 0. What you found probably did it that way because most calculators have a "tan" key but not a "cot" key (to find cot θ on such a calculator, enter θ, push the "tan" key, then push the "1/x" key).

After dividing across by cos θ, I ended up with the following trigonometric equation:

tan θ + 1 = 0

tan θ = -1

I am stuck here.
Really? If you enter "-1" on a calculator then take the "inverse tangent" (some calculator have a "second function" key that will give the inverse function) and you will get -0.78539816339744830961566084581988. Better would be to understand that tan θ= sin θ/cos θ so tan θ= -1 means that sin θ= -cos θ. You should know that sine and cosine are the same for \(\displaystyle \theta= \pi/4\) and that the signs of (sine, cosine) are (+, +) in the first quadrant, (+, -) in the second quadrant, (-, -) in the third quadrant and (-, +) in the second quadrant. So tan θ is negative in the second and fourth quadrants. That is, for θ between 0 and \(\displaystyle 2\pi\), tan θ= -1 for \(\displaystyle x= \pi/2+ \pi/4= 3\pi/4\) and \(\displaystyle x= 3\pi/2+ \pi/4= 7\pi/4\). If you want the result in degrees, those are 90+ 45= 135 degrees and 270+ 45= 315 degrees. You can get all solutions by adding integer multiples of \(\displaystyle 2
\pi\) (360 degrees) to those.
 
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skeeter

MHF Helper
Jun 2008
16,217
6,765
North Texas
for $0 \le \theta \le 2\pi$, $\tan{\theta} = -1$ at $\theta = \dfrac{3\pi}{4}$ and at $\theta = \dfrac{7\pi}{4}$

note that $\tan{\theta} = -1 \implies \cos{\theta} = -\sin{\theta}$

learn the unit circle ...

 
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May 2016
368
5
NYC
What is the value of r in terms of each point (r, θ)?
 

skeeter

MHF Helper
Jun 2008
16,217
6,765
North Texas
What is the value of r in terms of each point (r, θ)?
$r= 1+\cos\left(\dfrac{3\pi}{4}\right) = 1-\dfrac{\sqrt{2}}{2}$

you evaluate $r$ for the other point of intersection
 
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May 2016
368
5
NYC
$r= 1+\cos\left(\dfrac{3\pi}{4}\right) = 1-\dfrac{\sqrt{2}}{2}$

you evaluate $r$ for the other point of intersection
This is exactly what I thought in terms of finding r but I needed to sure.