Intersection pts of polar equations

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SUMMARY

The discussion focuses on finding the area of the region enclosed by the polar curves r = sin(θ) and r = sin(2θ). The key step involves determining the points of intersection by setting sin(θ) equal to sin(2θ). The critical intersection point is established at θ = π/3, derived from the equation cos(θ) = 1/2. This intersection point is essential for defining the limits of integration necessary for calculating the area.

PREREQUISITES
  • Understanding of polar coordinates and equations
  • Knowledge of trigonometric identities, specifically sin(2θ) = 2sin(θ)cos(θ)
  • Familiarity with the concept of integration in calculus
  • Ability to solve equations involving arcsin and trigonometric functions
NEXT STEPS
  • Study the derivation of polar area formulas, specifically for r = f(θ)
  • Learn about the graphical representation of polar equations and their intersections
  • Explore advanced trigonometric identities and their applications in solving equations
  • Practice integration techniques for polar coordinates, focusing on area calculations
USEFUL FOR

Students studying calculus, particularly those focused on polar coordinates, as well as educators and tutors looking to enhance their understanding of trigonometric intersections and area calculations in polar systems.

n00neimp0rtnt
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Homework Statement


I have to find the area of the region that lies inside the curves:

r = sin(θ)
r = sin(2θ)


The Attempt at a Solution



I'm assuming the first step would be to find the points of intersection so I know WHERE to integrate from/to, so I set the equations equal to each other:

sin(θ) = sin(2θ)


arcsin both sides:
θ = 2θ

And I'm stuck. Analysis of the graph shows that the most crucial intersection point occurs at or very close to 75º, but I would like to be able to show that.
 
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Taking arcsin of both sides will only give you some solutions. Try using sin(2x)=2*sin(x)*cos(x).
 
Thanks a lot! In that case...

sin(θ) = 2sin(θ)cos(θ)
1 = 2cos(θ)
cos(θ) = 1/2
θ = π/3

That should help me get the rest of the problem, thanks again! =]
 

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