tangent question: What is value of 1/tan(pi*x) when x = 1/2? Is it 0?

Harry_the_cat

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What is the value of 1tan(πx)\displaystyle \frac{1}{\tan(\pi x)} when x=12\displaystyle x = \frac{1}{2} ? Is it 0?

I know that tanθ=sinθcosθ\displaystyle \tan \theta = \frac{\sin \theta}{\cos \theta} tells us it is 0, but does that definition hold if cosθ=0\displaystyle \cos \theta =0.
 
yes …

cos(π2)sin(π2)=01=0\dfrac{\cos\left(\frac{\pi}{2}\right)}{\sin\left(\frac{\pi}{2}\right)} = \dfrac{0}{1} = 0
 
yes …

cos(π2)sin(π2)=01=0\dfrac{\cos\left(\frac{\pi}{2}\right)}{\sin\left(\frac{\pi}{2}\right)} = \dfrac{0}{1} = 0
Yes I can see that, but what happens when you plug x=12\displaystyle x=\frac{1}{2} into tan(πx)\displaystyle \tan(\pi x) ?
 
What is the value of 1tan(πx)\displaystyle \frac{1}{\tan(\pi x)} when x=12\displaystyle x = \frac{1}{2} ? Is it 0?

I know that tanθ=sinθcosθ\displaystyle \tan \theta = \frac{\sin \theta}{\cos \theta} tells us it is 0, but does that definition hold if cosθ=0\displaystyle \cos \theta =0.
cot(π/2)=cos(π/2)sin(π/2)=01=0cot(\pi/2) = \dfrac{cos(\pi/2)}{sin(\pi/2)} = \dfrac{0}{1} = 0

But
1tan(π/2)=  1  sin(π/2)cos(π/2)=  1  10\dfrac{1}{tan(\pi/2)} = \dfrac{ \, \, 1\,\, }{ \dfrac{sin(\pi/2)}{cos(\pi/2)} } = \dfrac{ \, \, 1\,\, }{ \dfrac{1}{0} }
does not exist.

-Dan
 
Do those little grey dots on the x-intercepts on Desmos represent discontinuities?
 
Do those little grey dots on the x-intercepts on Desmos represent discontinuities?
They should, but unfortunately they represent intersections. (There are no dots when this equation isn't selected.) Both Desmos and Wolfram Alpha apparently don't recognize the issue.

Similarly, I've seen sources that define cot(x) as 1/tan(x), which is wrong.
 
They should, but unfortunately they represent intersections. (There are no dots when this equation isn't selected.) Both Desmos and Wolfram Alpha apparently don't recognize the issue.

Similarly, I've seen sources that define cot(x) as 1/tan(x), which is wrong.
and also tan(x)\displaystyle \tan(x) as sin(x)cos(x)\displaystyle \frac{\sin(x)}{\cos(x)} ??
 
They should, but unfortunately they represent intersections. (There are no dots when this equation isn't selected.) Both Desmos and Wolfram Alpha apparently don't recognize the issue.

Similarly, I've seen sources that define cot(x) as 1/tan(x), which is wrong.
First time I’ve heard that … how should cotangent be defined, then?
 
and also tan(x)\displaystyle \tan(x) as sin(x)cos(x)\displaystyle \frac{\sin(x)}{\cos(x)} ??
No, that's perfectly valid, as is cot(x)=cos(x)sin(x)\displaystyle \cot(x)=\frac{\cos(x)}{\sin(x)}.

It can also be properly defined as x/y on the unit circle, or as tan(πx)\tan(\pi-x). All of these have the correct domain.

It is only valid to define cot(x)=1tan(x)\displaystyle \cot(x)=\frac{1}{\tan(x)} if you specify that you are working in the projectively extended reals, as I've seen stated in one place. Over the reals, it gets the wrong domain.

I've been looking for the sources I'd seen that gave that latter equation as a definition (rather than an identity that is valid when both sides are defined), but can't currently find it except in one dictionary. I suspect it is not uncommon for students to read the identity and think of it as a definition.
 
discussion at stack exchange ...

 
A necessary condition for a fraction to equal 0 is if the numerator is 0. The op's numerator is 1 which is never 0.
 
It is only valid to define cot(x)=1tan(x)\displaystyle \cot(x)=\frac{1}{\tan(x)} if you specify that you are working in the projectively extended reals, as I've seen stated in one place. Over the reals, it gets the wrong domain.
Can i check what you mean by 'projectively extended reals'? Is this the real line but also includes infinity somehow?
 
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