I want to prove the asymptotes for the inverse cotangent.

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SUMMARY

The discussion centers on proving the horizontal asymptotes for the inverse cotangent function, arccot x. It is established that the asymptotes are y = 0 and y = π, as arccot x cannot equal these values. The relationship arccot x = (π/2) - arctan x is critical, demonstrating that arctan x cannot approach ±π/2, which would otherwise yield the asymptotes. The discussion concludes with a suggestion to reflect the graph of y = cot(x) across the line y = x to visualize the behavior of y = arccot(x).

PREREQUISITES
  • Understanding of inverse trigonometric functions, specifically arccotangent.
  • Familiarity with the properties of arctangent and its asymptotic behavior.
  • Knowledge of horizontal asymptotes in the context of function limits.
  • Basic graphing skills to visualize transformations of trigonometric functions.
NEXT STEPS
  • Study the properties of inverse trigonometric functions, focusing on arccotangent.
  • Learn about horizontal asymptotes and how to determine them for various functions.
  • Explore the relationship between cotangent and its inverse, including graphical transformations.
  • Investigate the limits of arctan x as x approaches positive and negative infinity.
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Students studying precalculus or calculus, educators teaching inverse trigonometric functions, and anyone interested in understanding the behavior of arccotangent and its asymptotes.

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Homework Statement


arccot x = (π/2) - arctan x
arccot x =/= π
arccot x =/= 0

Homework Equations


arccot x = 1/arctan x (if x > 0)
arccot x = 1/arctan x + π (if x < 0)
arccot x = π/2 (if x = 0)

The Attempt at a Solution


π and 0 are the horizontal asymptotes, the values for which y (sine) cannot be.

arccot x = (π/2) - arctan x
arccot x =/= π
π =/= (π/2) - arctan x
(π/2) =/= -arctan x
-(π/2) =/= arctan x

arccot x = (π/2) - arctan x
arccot x =/= 0
0 =/= (π/2) - arctan x
-(π/2) =/= -arctan x
(π/2) =/= arctan x

Therefore, arctan x cannot be equal to negative or positive π/2, because if it were, then adding it to the original equation would produce π or 0, respectively. This cannot work because these are the values that it cannot be. As I write this, I'm thinking that I'm only establishing the fact that 0 and π are the asymptotes. But how would I prove that they're the asymptotes? If it's beyond the scope of precalculus, please say so.
 
Last edited:
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Eclair_de_XII said:

Homework Statement


arccot x = (π/2) - arctan x
arccot x =/= π
arccot x =/= 0

Homework Equations


arccot x = 1/arctan x (if x > 0)
arccot x = 1/arctan x + π (if x < 0)
arccot x = π/2 (if x = 0)

The Attempt at a Solution


π and 0 are the horizontal asymptotes, the values for which y (sine) cannot be.

arccot x = (π/2) - arctan x
arccot x =/= π
π =/= (π/2) - arctan x
(π/2) =/= -arctan x
-(π/2) =/= arctan x

arccot x = (π/2) - arctan x
arccot x =/= 0
0 =/= (π/2) - arctan x
-(π/2) =/= -arctan x
(π/2) =/= arctan x

Therefore, arctan x cannot be equal to negative or positive π/2, because if it were, then adding it to the original equation would produce π or 0, respectively. This cannot work because these are the values that it cannot be. As I write this, I'm thinking that I'm only establishing the fact that 0 and π are the asymptotes. But how would I prove that they're the asymptotes? If it's beyond the scope of precalculus, please say so.
What is there to prove? You can reflect the graph of y = cot(x) across the line y = x to get the graph of y = arccot(x).
 
Last edited:

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