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

## 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?

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etotheipi
2020 Award
For what values of ##\alpha## does ##\tan{\alpha}## diverge to positive infinity?

ElectronicTeaCup
PeroK
Homework Helper
Gold Member
2020 Award
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 and etotheipi
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
Homework Helper
Gold Member
2020 Award
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°##.

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

etotheipi
2020 Award
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
Mentor
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.

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
etotheipi
2020 Award
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).

archaic
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.

ElectronicTeaCup
haruspex