Question

How does sin(t)/(1-cos(t)) simplify to cos(t)/sin(t)?

Answer 1

are you sure that's the exact question as written?

Answer 2

It doesn't. Pick t=5 and you'll see it doesn't work.

Answer 3

For example...

Answer 4

Well it's that I'm looking at a solution my teacher wrote. He was showing how a cycloid had infinitely sharp cusps by finding the limit of the derivative of the parametric formulas given. So he was trying first to find the limit of sin(t)/(1-cos(t)) and he jumped to trying to find the limit of cos(t)/sin(t) because apparently, they're equal.

Answer 5

Look at #5. http://people.sfcollege.edu/bruce.teague/pdf/2312%20W11/2312_MG5_solutions_W11.pdf
This link should work.

Answer 6

Did everybody leave?

Answer 7

Awww c'mon!

Answer 8

Now what am I gonna do?

Answer 9

Calm down...I'll have a look.

Answer 10

Haha ok

Answer 11

Oh, he's used L'Hopital's rule...do you know about that?

Answer 12

Maybe. I might have forgotten.

Answer 13

I was about say that

Answer 14

When you have indeterminate forms for the limit (numerator and denominator each go to something like 0/0 or infinity/infinity), the limit of the original ratio is equal to the limit ratio of the derivatives of the numerator and denominator.

Answer 15

Take the derivative of the numerator of sin(t) and you get cos(t), and the derivative of 1-cost(t) is sin(t).

Answer 16

Oh! It's all coming back now.

Answer 17

OK. That makes sense. Thank you soooooo much.

Answer 18

Fan me then! :) I want to reach superstar!

Answer 19

How does one fan someone else?

Answer 20

\[\lim_{t \rightarrow 0} \frac{sin(t)}{1-\cos(t)}= \frac{0}{1-1}=\frac{0}{0}\]
lokisan is right, this is an indeterminate and using L'hopital would give you

Answer 21

there should be a link-type thing next to my name..."Become a fan"

Answer 22

I'm not seeing it. I'm looking.

Answer 23

It's in the thread window...little 'thumbs up' icon. There's one next to your own name too.

Answer 24

There. I got it.

Answer 25

Awesome...cheers!