Relating Fractions on the Unit Circle

[Jason76]

\(\displaystyle -\dfrac{1}{45}\) should equal \(\displaystyle \dfrac{\pi}{4}\)

\(\displaystyle \dfrac{1}{45}\) should equal \(\displaystyle \dfrac{3\pi}{4}\)

But I think something is wrong here

Reasoning:

Given: \(\displaystyle y = 2\cot(x)\)

Considering: \(\displaystyle y = A\cot(Bx - C) + D\)

Looking for extra points between the \(\displaystyle x\) intercept and respective asymptotes.

Given: Asympotes of \(\displaystyle 0\) and \(\displaystyle \pi\) and \(\displaystyle x\) intercept of \(\displaystyle \dfrac{\pi}{2}\)

\(\displaystyle A = y\) coordinates with positive and negative versions of \(\displaystyle A\) which equal each respective coordinate. Now, finding respective \(\displaystyle x\) coordinates by plugging in respective \(\displaystyle y\) coordinates.


a).-

\(\displaystyle -2 = 2\cot x\)

\(\displaystyle -2 = 2 \dfrac{1}{\tan x}\)

\(\displaystyle - 1 = \dfrac{1}{\tan x}\)

\(\displaystyle \arctan(-1) = \arctan(\dfrac{1}{\tan x})\)

\(\displaystyle -45 = \dfrac{1}{x}\)

\(\displaystyle x = -\dfrac{1}{45}\)

b).-

\(\displaystyle 2 = 2\cot x\)

\(\displaystyle 2 = 2 \dfrac{1}{\tan x}\)

\(\displaystyle 1 = \dfrac{1}{\tan x}\)

\(\displaystyle \arctan(1) = \arctan(\dfrac{1}{\tan x})\)

\(\displaystyle 45 = \dfrac{1}{x}\)

\(\displaystyle x = \dfrac{1}{45}\)

Now we can see that our two extra points are: \(\displaystyle (-\dfrac{1}{45},-1)\) and \(\displaystyle (\dfrac{1}{45},1)\)- maybe

But we know for sure the two points are : \(\displaystyle (\dfrac{\pi}{4},-1)\) and \(\displaystyle (\dfrac{3\pi}{4},1)\)-

Note: You could not have found the \(\displaystyle x\) intercept using inverse trig functions cause it sometimes leads to an undefined answer.

Of course, I know there is an easier way to find the extra points, but still want to be able to understand relation between the 1 over 45 degree fractions in question and \(\displaystyle \dfrac{\pi}{4}\) and \(\displaystyle \dfrac{3\pi}{4}\)

 

[Deleted member 4993]

\(\displaystyle -\dfrac{1}{45}\) should equal \(\displaystyle \dfrac{\pi}{4}\) ............... Absolutely not (not in this universe!!)

However you can say:

\(\displaystyle {45°}\) should equal \(\displaystyle \dfrac{\pi}{4} \ \ radians\)

\(\displaystyle \dfrac{1}{45}\) should equal \(\displaystyle \dfrac{3\pi}{4}\)

But I think something is wrong here

Reasoning:

Given: \(\displaystyle y = 2\cot(x)\)

Considering: \(\displaystyle y = A\cot(Bx - C) + D\)

Looking for extra points between the \(\displaystyle x\) intercept and respective asymptotes.

Given: Asympotes of \(\displaystyle 0\) and \(\displaystyle \pi\) and \(\displaystyle x\) intercept of \(\displaystyle \dfrac{\pi}{2}\)

\(\displaystyle A = y\) coordinates with positive and negative versions of \(\displaystyle A\) which equal each respective coordinate. Now, finding respective \(\displaystyle x\) coordinates by plugging in respective \(\displaystyle y\) coordinates.


a).-

\(\displaystyle -2 = 2\cot x\)

\(\displaystyle -2 = 2 \dfrac{1}{\tan x}\)

\(\displaystyle - 1 = \dfrac{1}{\tan x}\)

\(\displaystyle \arctan(-1) = \arctan(\dfrac{1}{\tan x})\)

\(\displaystyle -45 = \dfrac{1}{x}\)

\(\displaystyle x = -\dfrac{1}{45}\)

b).-

\(\displaystyle 2 = 2\cot x\)

\(\displaystyle 2 = 2 \dfrac{1}{\tan x}\)

\(\displaystyle 1 = \dfrac{1}{\tan x}\)

\(\displaystyle \arctan(1) = \arctan(\dfrac{1}{\tan x})\)

\(\displaystyle 45 = \dfrac{1}{x}\)

\(\displaystyle x = \dfrac{1}{45}\)

Now we can see that our two extra points are: \(\displaystyle (-\dfrac{1}{45},-1)\) and \(\displaystyle (\dfrac{1}{45},1)\)- maybe

But we know for sure the two points are : \(\displaystyle (\dfrac{\pi}{4},-1)\) and \(\displaystyle (\dfrac{3\pi}{4},1)\)-

Note: You could not have found the \(\displaystyle x\) intercept using inverse trig functions cause it sometimes leads to an undefined answer.

Of course, I know there is an easier way to find the extra points, but still want to be able to understand relation between the 1 over 45 degree fractions in question and \(\displaystyle \dfrac{\pi}{4}\) and \(\displaystyle \dfrac{3\pi}{4}\)

.

 

[mmm4444bot]

\(\displaystyle -\dfrac{1}{45}\) should equal \(\displaystyle \dfrac{\pi}{4}\)

But I think something is wrong here

You meant to type -Pi/4 above, yes? You know that a negative number can never equal a positive number.

You're mixing degree and radian measure together; that won't work in equations.

How about using radians? There is stuff in calculus that won't work with degree measure, so you need to be ready.

\(\displaystyle \arctan(\dfrac{1}{\tan x})\)

\(\displaystyle \dfrac{1}{x}\)

This result is not true, in general. (It works when x = 1 because the reciprocal of 1 is itself.)

For example, let x = Pi/6, and you'll see that arctan(1/tan[Pi/6]) is not 6/Pi.



Back up to your prior result:

-1 = 1/tan(x)

At this point, I would solve the equation for tan(x).

tan(x) = -1

Now take the arctangent of both sides (and please use radians) :cool:

 

[Jason76]

I see what's going on now. \(\displaystyle -\dfrac{1}{45}\) degrees is equal to \(\displaystyle 45\) degrees, and \(\displaystyle \dfrac{1}{45}\) degrees is equal to \(\displaystyle -45\) degrees. Note that \(\displaystyle -45\) degrees is the same as \(\displaystyle \dfrac{3\pi}{4}\)

Most of this stuff is true because: \(\displaystyle -\dfrac{1}{a} = a\) and \(\displaystyle \dfrac{1}{a} = -a\)

 

[Deleted member 4993]

I see what's going on now.

\(\displaystyle -\dfrac{1}{45}\) degrees is equal to \(\displaystyle 45\) degrees, ...... Incorrect

and \(\displaystyle \dfrac{1}{45}\) degrees is equal to \(\displaystyle -45\) degrees. ...... Incorrect


Note that

\(\displaystyle -45\) degrees is the same as \(\displaystyle \dfrac{3\pi}{4}\)...... Incorrect

Most of this stuff is true because:

\(\displaystyle -\dfrac{1}{a} = a\) ...... Incorrect

and

\(\displaystyle \dfrac{1}{a} = -a\)...... Incorrect

.

 

[Jason76]

I got this all wrong. You can't use inverse trig functions to find the extra points. For example, if you do the "inverse trig function way" for the extra points, then you get \(\displaystyle -\dfrac{\pi}{4}\) However, that's NOT between the asymptotes of \(\displaystyle 0\) and \(\displaystyle \pi\) Nonetheless, the other answer of \(\displaystyle \dfrac{\pi}{4}\) is correct.

I don't know the mathematical reason for all of this, but I guess it's not important at this stage.

Anyhow, the correct way to find the extra points involves multiplying the total area by \(\displaystyle \dfrac{1}{4}\) and \(\displaystyle \dfrac{3}{4}\) respectively to get the center \(\displaystyle x\) coordinate for each respective side (left and right).

 

[MarkFL]

I got this all wrong. You can't use inverse trig functions to find the extra points. For example, if you do the "inverse trig function way" for the extra points, then you get \(\displaystyle -\dfrac{\pi}{4}\) However, that's NOT between the asymptotes of \(\displaystyle 0\) and \(\displaystyle \pi\) Nonetheless, the other answer of \(\displaystyle \dfrac{\pi}{4}\) is correct.

I don't know the mathematical reason for all of this, but I guess it's not important at this stage.

Anyhow, the correct way to find the extra points involves multiplying the total area by \(\displaystyle \dfrac{1}{4}\) and \(\displaystyle \dfrac{3}{4}\) respectively to get the center \(\displaystyle x\) coordinate for each respective side (left and right).

Without knowing to what you are referring, I have no idea what you are talking about. It looks like you are thinking out loud about a problem, the details of which you have not shared. How are we to provide any meaningful input?

 

[Jason76]

Without knowing to what you are referring, I have no idea what you are talking about. It looks like you are thinking out loud about a problem, the details of which you have not shared. How are we to provide any meaningful input?

Graphing \(\displaystyle y = 2 \cot(x)\)

Considering: \(\displaystyle y = A\cot(Bx - C) + D\)

Assymptotes are \(\displaystyle 0\) and \(\displaystyle \pi\) - Found by setting \(\displaystyle Bx - C\) (which in this case is \(\displaystyle 1\)) to \(\displaystyle 0\) and \(\displaystyle \pi\) respectively and solving for \(\displaystyle x\).

Area between asymptotes is \(\displaystyle \pi\) - Is there a formula to find the area between asymptotes?

\(\displaystyle x\) intercept is \(\displaystyle \dfrac{\pi}{2}\) - Found by multiplying the area between the asymptotes by \(\displaystyle \dfrac{1}{2}\)

Find the extra points between each respective asymptote and the \(\displaystyle x\) intercept. - The key is multiplying the area by \(\displaystyle \dfrac{1}{4}\) to get the \(\displaystyle x\) coordinate of the 1st point (to the left of the \(\displaystyle x\) intercept), and then multiplying the area by \(\displaystyle \dfrac{3}{4}\) to get the \(\displaystyle x\) coordinate of the 2nd point (to the right of the \(\displaystyle x\) intercept). Now, find \(\displaystyle y\) coordinates of the extra points by plugging in respective \(\displaystyle x\) coordinates and solving for \(\displaystyle y\) respectively. two points are : \(\displaystyle (\dfrac{\pi}{4}, -2)\) and \(\displaystyle (\dfrac{3\pi}{4}, 2)\) Note the \(\displaystyle y\) coordinates correspond to positive and negative versions of \(\displaystyle A\) (amplitude) -

Finally, connect the dots to graph.

Note: I did try to use inverse trig functions to get answers, but it doesn't work.

 

[Deleted member 4993]

Graphing \(\displaystyle y = 2 \cot(x)\)

Considering: \(\displaystyle y = A\cot(Bx - C) + D\)

Assymptotes are \(\displaystyle 0\) and \(\displaystyle \pi\) - Found by setting \(\displaystyle Bx - C\) (which in this case is \(\displaystyle 1\)) to \(\displaystyle 0\) and \(\displaystyle \pi\) respectively and solving for \(\displaystyle x\).

Area between asymptotes is \(\displaystyle \pi\) - Is there a formula to find the area between asymptotes?

\(\displaystyle x\) intercept is \(\displaystyle \dfrac{\pi}{2}\) - Found by multiplying the area between the asymptotes by \(\displaystyle \dfrac{1}{2}\)

Find the extra points between each respective asymptote and the \(\displaystyle x\) intercept. - The key is multiplying the area by \(\displaystyle \dfrac{1}{4}\) to get the \(\displaystyle x\) coordinate of the 1st point (to the left of the \(\displaystyle x\) intercept), and then multiplying the area by \(\displaystyle \dfrac{3}{4}\) to get the \(\displaystyle x\) coordinate of the 2nd point (to the right of the \(\displaystyle x\) intercept). Now, find \(\displaystyle y\) coordinates of the extra points by plugging in respective \(\displaystyle x\) coordinates and solving for \(\displaystyle y\) respectively. two points are : \(\displaystyle (\dfrac{\pi}{4}, -2)\) and \(\displaystyle (\dfrac{3\pi}{4}, 2)\) Note the \(\displaystyle y\) coordinates correspond to positive and negative versions of \(\displaystyle A\) (amplitude) -

Finally, connect the dots to graph.

Note: I did try to use inverse trig functions to get answers, but it doesn't work.

You have plotted y = tan(x)

just use:

cot(Θ) = tan(π/2 - Θ)

 

[Jason76]

Again Doing This All Wrong. Here is the right way:

\(\displaystyle y = 2\cot x\)

Given formula: \(\displaystyle y = A\cot(Bx - C) + D\)

1. Identify the Letters

\(\displaystyle A = 2, B = 1, C = 0, D = 0\)

2. Locate Vertical Asymptotes

Set \(\displaystyle Bx - C\) respectively to \(\displaystyle 0\) and \(\displaystyle \pi\)

a.)-

\(\displaystyle x - 0 = 0\)

\(\displaystyle x = 0\)

b).-

\(\displaystyle x - 0 = \pi\)

\(\displaystyle x = \pi\)

So our two vertical asymptotes are \(\displaystyle 0\) and \(\displaystyle \pi\)

3. Calculate the Area

right asymptote \(\displaystyle -\) left asymptote\(\displaystyle =\) area

\(\displaystyle \pi - 0 = \pi\)

\(\displaystyle area = \pi\)

4. Identify the x intercept

Simply multiply the area by \(\displaystyle \dfrac{1}{2}\). Next, add the answer to the left asympotote.

\(\displaystyle [(\dfrac{1}{2})\)area\(\displaystyle ] +\) left asymptote\(\displaystyle = x\) intercept.

\(\displaystyle [(\dfrac{1}{2})\pi] + 0 = 0\)

The \(\displaystyle x\) intercept is \(\displaystyle \dfrac{\pi}{2}\)

5. Find Two More Points

In order to make a more accurate graph, we need two points. These points should be located between the \(\displaystyle x\) axis and each respective asympote.

Left Point \(\displaystyle x\) coordinate \(\displaystyle = [(\dfrac{1}{4})\)area\(\displaystyle ] + \)left asymptote.

Left Point \(\displaystyle x\) coordinate \(\displaystyle = [(\dfrac{1}{4})\pi] + 0 = \dfrac{\pi}{4}\):

Right Point \(\displaystyle x\) coordinate \(\displaystyle = [(\dfrac{3}{4})\)area\(\displaystyle ] + \) left asymptote.

Right Point \(\displaystyle x\) coordinate \(\displaystyle = [(\dfrac{3}{4})\pi] + 0 = \dfrac{3\pi}{4}\):

To find the respective \(\displaystyle y\) coordinates simply plug the respective \(\displaystyle x\) values into the original equation.

6. Graph the Function

Connect the dots.-

Feel free to correct. The method for graphing a tangent is the same in regards to the \(\displaystyle x\) intercept and extra points.