- #1
ElectronicTeaCup
- 23
- 1
- 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 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?
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}##
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.archaic said:Isn't ##1-1## a bit ambiguous, though? Was this a limit like ##x-1##?
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##).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##?
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.
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.ElectronicTeaCup said:Homework Statement:: Tan ##2 \theta=4 /(1-1)##.
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 said:I seem not to understand this problem
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.
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.
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.
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.
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.