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Base on the unit circle, I know

$ \begin{align} &\cot\left(\frac{\pi}{2}\right) \\ =&\frac{0}{1}\\ =&0 \end{align} $

But it is also

$ \begin{align} &\cot\left(\frac{\pi}{2}\right) \\ =&\frac{1}{\tan\left(\frac{\pi}{2}\right)}\\ =&\frac{1}{\frac{1}{0}}\\ =&undefined \end{align} $

And Google gives me this answer:

 $6.12303177 × 10^{-17}$

I am really confused now. Although I know it is $0$, I don't see why the other ones are wrong.

  • 0
    Yes, that tells you the only sensible interpretation of $1/\infty$ is zero. But if you want to be sure I see a question you ask me, you have to include @Gerry in it.2012-10-15

2 Answers 2

14

$\cot x = \frac{1}{\tan x}$ only when $\tan x \neq 0$ (i.e. $x \neq n\pi$ for any $n\in \mathbb {Z}$).

However, $\cot x$ is actually defined as

$\cot x := \frac{\cos x}{\sin x}$ so $\cot \left(\frac{\pi}{2}\right)=0$ is the correct answer.

  • 10
    To add to Argon's answer, the Google answer is due to numerical round-off error.2012-10-14
-3

$\tan \frac π2$ is actually infinity.

So if we divide 1 by $\tan \frac π2$ we will have $\frac 1{\infty}$ which leads to 0.

$\cot \frac π2 = \frac 1{\tan \frac π2}=0$

Furthermore:

$\cot x=\frac 1{\tan x}=\frac{\cos x}{\sin x}$

So:

$\cot \frac π2=\frac{\cos \frac π2}{\sin \frac π2} = \frac 01 = 0$

  • 0
    We can't 'divide' a number by $\tan (\pi/2)$; $\tan(x)$ is undefined at $x=\pi/2$, unless we edit the definition of tangent function.2015-11-03