Author Topic: cot(x)=0 at x=pi/2? (Read 962 times) Tweet Share

horizon · « **on:** April 22, 2012, 08:25:32 pm »

If cot(x)=1/tan(x), then how can cot(x)=0 at x=pi/2?
Isn't tan(pi/2) undefined?

TrueTears · « **Reply #1 on:** April 22, 2012, 08:30:35 pm »

it's not a direct substitution $\lim_{x \to \frac{\pi}{2}^-} \tan(x) = +\infty \implies \lim_{x \to \frac{\pi}{2}^-} \cot(x) = 0$

Lasercookie · « **Reply #2 on:** April 22, 2012, 08:31:30 pm »

Well consider why tan(pi/2) is undefined:

$tan(x)=\frac{sin(x)}{cos(x)}$ , if $x = \frac{\pi}{2}$ and you'll be dividing by zero right? since $cos(\frac{\pi}{2}) = 0$

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

Well now it's flipped and you're no longer dividing by zero since $sin(\frac{\pi}{2})=1$ , and it's all good.

brightsky · « **Reply #3 on:** April 22, 2012, 08:41:42 pm »

Quote from: TrueTears on April 22, 2012, 08:30:35 pm

it's not a direct substitution $\lim_{x \to \frac{\pi}{2}^-} \tan(x) = +\infty \implies \lim_{x \to \frac{\pi}{2}^-} \cot(x) = 0$

how would you prove continuity at x = pi/2?

TrueTears · « **Reply #4 on:** April 22, 2012, 08:49:37 pm »

then we have to show that $\lim_{x \to \frac{\pi}{2}}{\cot(x)} = \cot(\frac{\pi}{2})$

which is true since $\lim_{x \to \frac{\pi}{2}}{\frac{\cos(x)}{\sin(x)}} = \frac{\cos(\frac{\pi}{2})}{\sin(\frac{\pi}{2})}$

Special At Specialist · « **Reply #5 on:** April 25, 2012, 02:01:13 pm »

tan(pi/2) = 1/0 = undefined
cot(pi/2) = 0/1 = 0

Search the forums now!

We have moved!

Author Topic: cot(x)=0 at x=pi/2? (Read 962 times) Tweet Share

horizon

cot(x)=0 at x=pi/2?

TrueTears

Re: cot(x)=0 at x=pi/2?

Lasercookie

Re: cot(x)=0 at x=pi/2?

brightsky

Re: cot(x)=0 at x=pi/2?

TrueTears

Re: cot(x)=0 at x=pi/2?

Special At Specialist

Re: cot(x)=0 at x=pi/2?

Recent Posts

User:
Password: