Determining 4 intersection points of two polar graphs?

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SUMMARY

The discussion focuses on finding the four intersection points of the polar graphs defined by the equations r = 2 and r^2 = 9sin(2θ). By equating the two equations, the solution reveals that 4/9 = sin(2θ). The inverse sine function provides one solution, while the general solution for θ is derived using the arcsine function and periodicity. The four specific angles are θ = (1/2)arcsin(4/9), (π/2) - (1/2)arcsin(4/9), π + (1/2)arcsin(4/9), and (3π/2) - (1/2)arcsin(4/9).

PREREQUISITES
  • Understanding of polar coordinates and graphs
  • Knowledge of trigonometric identities, specifically sin(2θ)
  • Familiarity with the inverse sine function and its properties
  • Basic skills in solving equations involving trigonometric functions
NEXT STEPS
  • Study the properties of polar coordinates and their graphical representations
  • Learn about trigonometric identities, particularly double angle formulas
  • Explore the concept of periodicity in trigonometric functions
  • Practice solving polar equations and finding intersection points
USEFUL FOR

Students studying trigonometry, mathematics educators, and anyone interested in solving polar equations and understanding their graphical intersections.

Tekee
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Homework Statement


r = 2
r^2 = 9sin(2theta)
Find the 4 points of intersection


Homework Equations





The Attempt at a Solution


Since r = 2, 4 = 9sin(2theta)...
4/9 = sin(2theta)

Taking the inverse of 4/9 only gives me one answer on the calculator (obviously), and I do not know where to attain the 3 other points. (also, is there a way to do this without a calculator/calculus?) - IE 4/9 = 2sin(theta)cos(theta), although I do not see how this helps.
 
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[tex]\sin 2\theta=\frac{4}{9}\Rightarrow 2\theta =(-1)^k\arcsin\frac{4}{9}+k\pi\Rightarrow\theta =(-1)^k\frac{\arcsin\frac{4}{9}}{2}+\frac{k\pi}{2}[/tex]
[tex]k=0\Rightarrow\theta=\frac{1}{2}\arcsin\frac{4}{9}[/tex]
[tex]k=1\Rightarrow\theta=\frac{\pi}{2}-\frac{1}{2}\arcsin\frac{4}{9}[/tex]
[tex]k=2\Rightarrow\theta=\pi+\frac{1}{2}\arcsin\frac{4}{9}[/tex]
[tex]k=3\Rightarrow\theta=\frac{3\pi}{2}-\frac{1}{2}\arcsin\frac{4}{9}[/tex]
 

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