2
$\begingroup$

Some time ago I was wondering about the definition of cotangent; namely, that it is both defined as $\frac{\cos x}{\sin x}$ and as $\frac{1}{\tan x}$. However, $\cot(90°)$ and $\cot(270°)$ are equal to 0. Through the first definition I listed of cotangent, this is equal to $$\cot(90°) =\frac{\cos(90°)}{\sin(90°)} = \frac{0}{1} = 0$$

With the second definition, this is equal to $$\cot(90°) = \frac{1}{\tan(90°)} = \frac{1}{undefined} =\ ?$$

While I know that at $\tan(90°)$ there is an asymptote which, from the right goes to $-\infty$ and from the left goes to $\infty$. While I understand that in this case, the limit $\lim_{x\to \frac{\pi}{2}}\frac{1}{\tan x} = 0$, wouldn't this constitute a domain error in the function $\cot x$ since $\frac{1}{\infty}$ technically doesn't equal $0$?

Thanks for reading.

$\endgroup$
  • $\begingroup$ It is no different as defining $f(x) = x$ as $\frac{1}{1/x}$. Depends on definition i think. $\endgroup$ – King Tut Feb 1 '18 at 17:31
3
$\begingroup$

Cotangent is not $\frac 1\tan$. That is not the definition, and never has been. It just so happens that the two functions $\cot$ and $\frac1\tan$ agree on almost all of $\Bbb R$. When that happens, there is a strict sense in which we consider the functions equivalent, and some authors misuse the symbol $=$ to represent this.

That being said, kudos to you for stopping up and thinking about it.

$\endgroup$
  • $\begingroup$ Then is the exact definition of cotangent $\frac {\cos x}{\sin x}$? $\endgroup$ – Gil Keidar Feb 1 '18 at 17:37
  • $\begingroup$ @GilKeidar Yes, that's right. $\endgroup$ – Arthur Feb 1 '18 at 17:37
1
$\begingroup$

Just $$\cot{x}\neq\frac{1}{\tan{x}}$$ on domain of $\cot.$

We can write $\cot{x}=\frac{1}{\tan{x}}$ for $\sin{x}\cos{x}\neq0$ only.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.