Tan ##2 \theta=4 /(1-1)##. This means ##2 \theta=90^{\circ}## Why?

[ElectronicTeaCup]

Homework Statement

Tan $2 \theta=4 /(1-1)$. This means $2 \theta=90^{\circ}$ and $\theta=45^{\circ}$

Relevant Equations

None

One of my solutions had this in one part. Why is this the case?

 

[etotheipi]

For what values of $\alpha$ does $\tan{\alpha}$ diverge to positive infinity?

 

[PeroK]

Homework Statement:: Tan $2 \theta=4 /(1-1)$. This means $2 \theta=90^{\circ}$ and $\theta=45^{\circ}$
Relevant Equations:: None

One of my solutions had this in one part. Why is this the case?

You know that as $\theta \rightarrow \frac \pi 2$ then $\tan \theta \rightarrow +\infty$?

 

[ElectronicTeaCup]

Oh right, I wasn't even thinking about infinity, I was just thinking of it as "undefined"

Also, is this also correct?


$\begin{array}{l} \cot 2 \theta=0 \\ \frac{\cos 2 \theta}{\sin 2 \theta}=0 \\ \cos 2 \theta=0 \\ 2 \theta=90 \end{array}$

 

[PeroK]

Oh right, I wasn't even thinking about infinity, I was just thinking of it as "undefined"

Also, is this also correct?


$\begin{array}{l} \cot 2 \theta=0 \\ \frac{\cos 2 \theta}{\sin 2 \theta}=0 \\ \cos 2 \theta=0 \\ 2 \theta=90 \end{array}$

That's right where $2\theta$ is between $0$ and $180°$. Another solution could be $2\theta = 270°$.

 

[archaic]

Isn't $1-1$ a bit ambiguous, though? Was this a limit like $x-1$?

 

[etotheipi]

Isn't $1-1$ a bit ambiguous, though? Was this a limit like $x-1$?

A related example... if we parameterise a circle with $(x,y) = (a\cos{\theta}, a\sin{\theta})$ s.t. $y' = -\frac{\cos{\theta}}{\sin{\theta}}$ and wanted to see where the circle is vertical, would you also take issue with identifying $\sin{\theta} = 0 \implies \theta = 0, \pi$?

 

[Mark44]

Isn't $1-1$ a bit ambiguous, though? Was this a limit like $x-1$?

1 - 1 = 0, which is not at all ambiguous. However, the fraction $\frac 4 {1 - 1}$ is undefined.

 

[archaic]

A related example... if we parameterise a circle with $(x,y) = (a\cos{\theta}, a\sin{\theta})$ s.t. $y' = -\frac{\cos{\theta}}{\sin{\theta}}$ and wanted to see where the circle is vertical, would you also take issue with identifying $\sin{\theta} = 0 \implies \theta = 0, \pi$?

No problem with identifying the vertical part, but $y'$ either goes to infinity or negative infinity depending on how you choose to make $\theta$ go to either of those values (right or left of $0$ or $\pi$).
The same thing with the $1/(1-1)$ fraction. Whether you have it $1-x$ or $x-1$, then, however $x$ approaches $1$, you will get different results ($\pi/2$ or $-\pi/2$).
@Mark44 this is what I was aiming at.

 

[etotheipi]

No problem with identifying the vertical part, but $y'$ either goes to infinity or negative infinity depending on how you choose to make $\theta$ go to either of those values (right or left of $0$ or $\pi$).
The same thing with the $1/(1-1)$ fraction. Whether you have it $1-x$ or $x-1$, then, however $x$ approaches $1$, you will get different results ($\pi/2$ or $-\pi/2$).
@Mark44 this is what I was aiming at.

I see your point, and I the interpretation would be clearer from context.

In most cases where I've seen these sorts of steps, it hasn't mattered from which side you approach $x$ (e.g. for my circle example, a gradient of $\infty$ is equivalent to one of $-\infty$ for purposes of determining the vertical tangent).

 

[romsofia]

A hand-wavy algebraic solution:

$$\tan 2x = \frac{2 \tan x}{1-\tan^2 x} = \frac{4}{0} \rightarrow 0 = 4-4\tan^2 x \rightarrow \tan^2 x = 1$$

And from here the $\frac{\pi}{4}$ falls out "neatly", along with all the other solutions. But be careful, this is "illegal" math. However, you started off with a statement of $\frac{4}{0}$, I feel okay using hand-wavy tactics here.

 

[haruspex]

Homework Statement:: Tan $2 \theta=4 /(1-1)$.

Is that really the problem as given to you? I suggest that you had done some work to arrive at that. If so, please post the actual problem.

 

[chwala]

I seem not to understand this problem, well what i know is that,
$tan 2x$ = $\dfrac {2 tan x}{1-tan ^2 x}$ and given that (from OP)$tan 2θ$ = $\dfrac {4}{1-1}$ if indeed $2tan x=4$ then $⇒tan x=2$ which clearly contradicts with $tan^2 x=1$.

 

[pbuk]

I seem not to understand this problem

With 0 in the denominator the problem is not well defined so there is no point wasting your time trying to understand it, move on.