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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Specialist Mathematics => Topic started by: TrueTears on January 17, 2010, 12:02:05 am

Title: lol this is funny
Post by: TrueTears on January 17, 2010, 12:02:05 am: Solve $\frac{1}{\tan(x)} = 0$

Clearly something of the form $\frac{1}{u} = 0$ will not have solutions.

But...

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

$\cot(x) = \frac{\cos(x)}{\sin(x)} = 0$

Thus $x = \frac{\pi}{2}, \frac{3\pi}{2} \cdots$

:uglystupid2: :uglystupid2: :uglystupid2:

What do we define in these situations?
Title: Re: lol this is funny
Post by: Mao on January 17, 2010, 01:11:24 am: Quote from: TrueTears on January 17, 2010, 12:02:05 am
Clearly something of the form $\frac{1}{u} = 0$ will not have solutions.

For $u \in \mathbb{R}$ , yes, that would be true. If u happens to be 'infinity', or something ridiculous like that, it may well be possible that the statement is true.

When solving $\frac{1}{\tan(x)} = 0$ , remember the maximal domain is $x \neq n\pi + \frac{\pi}{2},\; n \in \mathbb{Z}$ [maximal domain of tan(x)], and also $x \neq n\pi, \; n\in \mathbb{Z}$ [cannot divide by 0]

You need to be very careful in the statement that $\frac{1}{\tan (x)} = \cot (x)$ , the domain of these two functions are not the same, hence in this situation, not applicable.

So is there a solution to this? Limits would say yes, domain would say no. Practicality would say 'I don't care, why does it matter'. :P

[To '10 specialist students, don't worry if you cannot understand this now, you will soon :)]
Title: Re: lol this is funny
Post by: TrueTears on January 17, 2010, 01:17:25 am: haha true.

In the case of the context where this question came from, I was actually sketching $\log_e(\sin(x))$ and wanted to find the turning points which lead to the equation $\frac{1}{\tan(x)}$ which in this case I guess there would be solutions.

But yes if it was just purely solving $\frac{1}{\tan(x)} = 0$ then the domain definition would suffice and there would be no solutions for $x \in \mathbb{R}$ :D :D :D :D :D
Title: Re: lol this is funny
Post by: Juddinator on January 17, 2010, 02:37:49 pm: [quote[To '10 specialist students, don't worry if you cannot understand this now, you will soon :)]
[/quote]
Good to know! :D
Title: Re: lol this is funny
Post by: NE2000 on January 18, 2010, 03:22:26 pm: Think this discussion is relevant: http://vcenotes.com/forum/index.php/topic,10730.0.html
Title: Re: lol this is funny
Post by: TrueTears on January 18, 2010, 03:31:27 pm: Yeah, it's pretty trivial actually, just different definitions. Whatever suits the concept, use that definition :)