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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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