Find the inverse of following function

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SUMMARY

The function p(x) = (cos(2πx), sin(2πx)) maps the half-open interval [0, 1) in R to the unit circle in R². Due to its periodic nature, the function is not one-to-one, thus lacking an inverse outside the specified interval. The range of p(x) is defined as {(x,y) | x² + y² = 1}, which serves as the domain for the inverse function. The inverse function is given by p⁻¹(x,y) = arccos(x)/(2π) for y ≥ 0 and 1 - arccos(x)/(2π) for y < 0.

PREREQUISITES
  • Understanding of trigonometric functions and their properties
  • Familiarity with the concept of inverse functions
  • Knowledge of the unit circle in R²
  • Basic calculus, specifically dealing with intervals and mappings
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  • Study the properties of periodic functions and their inverses
  • Learn about the unit circle and its applications in trigonometry
  • Explore the concept of mapping intervals to geometric shapes
  • Investigate the implications of non-one-to-one functions in calculus
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Mathematicians, students studying calculus and trigonometry, and anyone interested in understanding inverse functions and their applications in geometry.

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How do you find the inverse of following function?

p(x)=(cos2pix,sin2pix) where pi means 3.14etc NOT p times i.
 
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This is a function from R to R2? I.e. [itex]p(t)= (cos(2\pi x), sin(2\pi x)[/itex]. Then it maps the half open interval [0, 1) in R onto the unit circle in R2. At t=1 you start around the circle again so the function is not one-to-one and you do not have an inverse outside that interval. The range of that function is, of course, {(x,y)|x2+ y2= 1} and that must be the domain of the inverse function. On that domain, [itex]p^{-1}(x,y)= arccos(x)/(2\pi)[/itex] if y>= 0, [itex]1- arcos(x)/(2\pi)[/itex] if y< 0.
 

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