Just to prove them wrong: Trigonometry

[mooncrater]

Homework Statement

There is a part of solution of a question which says that:
$cot|cot^{-1}x|=cot cot^{-1}x=x$

Homework Equations

The Attempt at a Solution

If we put $x=-1$ in this equation then $cot^{-1}(-1)=-\pi /4$ which will have a modulus =$\pi /4$. And $cot \pi /4=1$ which is not equal to the $x$ we took. So this equation seems wrong to me. Am I correct here?

 

[Raghav Gupta]

You should know the domain and ranges.
What is the range for cot^-1x ?

 

[mooncrater]

You should know the domain and ranges.
What is the range for cot^-1x ?

Hmmmm... I think I have lost it... it's domain is $R$ and has a range $(0, \pi)$. So $cot^{-1}(-1)=3\pi /4$ not $-\pi /4$.
I forgot its range...
So now this equation is true for all values of $x$.
Thanks ($R\rightarrow R$ help)

 

[epenguin]

Homework Statement

There is a part of solution of a question which says that:
$cot|cot^{-1}x|=cot cot^{-1}x=x$

Homework Equations

The Attempt at a Solution

If we put $x=-1$ in this equation then $cot^{-1}(-1)=-\pi /4$ which will have a modulus =$\pi /4$. And $cot \pi /4=1$ which is not equal to the $x$ we took. So this equation seems wrong to me. Am I correct here?

If it is not equal to the x you took, it is equal to
| (the x you took) |
and there is a | | in the problem.

I think the answer you have been given is wrong and the answer is | x | .

You seem to have been given a few wrong answers and even wrong questions, or at least at the limit thereof.

 

[SammyS]

If it is not equal to the x you took, it is equal to
| (the x you took) |
and there is a | | in the problem.

I think the answer you have been given is wrong and the answer is | x | .

You seem to have been given a few wrong answers and even wrong questions, or at least at the limit thereof.

The standard range used for the arccotangent is (0, π). Thus the arccotangent function always returns a positive value.

So we have | cot^-1(x) | = cot^-1(x)

Added in Edit:
However, the range used for the arccotangent function is not universally established.