Just to prove them wrong: Trigonometry

1. May 4, 2015

mooncrater

1. The problem statement, all variables and given/known data
There is a part of solution of a question which says that:
$cot|cot^{-1}x|=cot cot^{-1}x=x$

2. Relevant equations

3. 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?

2. May 4, 2015

Raghav Gupta

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

3. May 4, 2015

mooncrater

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)

4. May 5, 2015

epenguin

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.

5. May 5, 2015

SammyS

Staff Emeritus
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)