I want to prove the asymptotes for the inverse cotangent.

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In summary: The horizontal asymptotes of y = cot(x) are the vertical asymptotes of y = arccot(x). Therefore, the asymptotes for arccot(x) are x = 0 and x = π. In summary, the horizontal asymptotes for arccot(x) are x = 0 and x = π, as they are the vertical asymptotes for cot(x).
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Eclair_de_XII
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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.
 
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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:

1. What are asymptotes?

Asymptotes are lines that a curve approaches but never touches, as it extends towards infinity.

2. What is the inverse cotangent function?

The inverse cotangent function, also known as arccotangent or arctan, is the inverse of the cotangent function. It takes a ratio of the adjacent and opposite sides of a right triangle and returns the angle measure in radians.

3. Why is it important to prove the asymptotes for the inverse cotangent function?

Proving the asymptotes for the inverse cotangent function is important in understanding the behavior of the function as the input values approach infinity or negative infinity. It also helps in graphing the function accurately.

4. What is the process for proving the asymptotes for the inverse cotangent function?

The process involves setting the input value of the function to infinity or negative infinity and solving for the corresponding output value. If the output value approaches a specific number or becomes undefined, then that number or value is an asymptote for the function. This is usually done using limits and trigonometric identities.

5. Are there any other methods for proving the asymptotes for the inverse cotangent function?

Yes, one can also use the properties of inverse functions to prove the asymptotes. For example, the inverse cotangent function has the same asymptotes as the cotangent function, but with their x and y coordinates flipped. This can also be used to determine the asymptotes for the inverse cotangent function.

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