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

• ElectronicTeaCup
In summary, the conversation discusses the issue of a fraction with a denominator of 0 and its implications in various equations, including the tangent function. One of the solutions had this issue and the participants discuss how it relates to infinity and undefined values. They also mention specific examples and their interpretations in terms of limits and approaching values. The conversation concludes with the suggestion to move on from trying to understand the problem due to its lack of proper definition.
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?

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

ElectronicTeaCup
ElectronicTeaCup said:
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}##

ElectronicTeaCup said:
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##?

archaic said:
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##?

archaic said:
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.

etotheipi said:
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
archaic said:
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
ElectronicTeaCup said:
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.

SammyS and archaic
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##.

Last edited:
chwala said:
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.

chwala

## 1. What does the equation "Tan 2θ = 4/(1-1)" mean?

The equation is a trigonometric identity which states that the tangent of twice an angle (2θ) is equal to 4 divided by the difference between 1 and 1. This can also be written as tan 2θ = undefined, since division by 0 is undefined.

## 2. How is this equation related to the value of 2θ being 90 degrees?

By substituting 90 degrees for 2θ in the equation, we get tan 90 = undefined, which is true since the tangent of 90 degrees is undefined. This means that the equation is satisfied when 2θ is equal to 90 degrees.

## 3. Why is the value of 2θ being 90 degrees significant in this equation?

The value of 2θ being 90 degrees is significant because it is the only value that satisfies the equation and makes it true. Any other value for 2θ would result in an undefined or false statement.

## 4. Can this equation be solved to find the value of θ?

No, this equation cannot be solved to find the value of θ. Since the equation is an identity, it is true for all values of θ, except for 2θ = 90 degrees which is the only solution. Therefore, there is no specific value for θ that can be found through solving this equation.

## 5. How is this equation used in science?

This equation is used in science, specifically in trigonometry and geometry, to understand the relationship between angles and their trigonometric functions. It can also be used to solve problems involving triangles and other geometric shapes. In addition, this equation is used in physics and engineering to calculate and analyze the motion of objects in circular or oscillatory motion.

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