Prove that one trig function equals another (too lengthy to include in title)

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Eigendorf

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Hello!

I'm trying to show that one function is equal to another but I'm not sure how to peform the operations to show that they are equal.

I'm fairly certain that the functions are equal because when I plug them both into my graphing calculator the results are the same.


This is part of a longer homework problem I'm working on and my answer is different from what is in the back of the book but the graphs of the function are the same.

Anyways the function that I'm trying to figure out how to convert to the answer in the back of the book is:

My Function:

\(\displaystyle -4\ln{\dfrac{4 + \sqrt{16 - x^2}}{x}}\)

The Books Function:

\(\displaystyle 4\ln{\dfrac{4 - \sqrt{16 - x^2}}{x}}\)


These appear to be equal when graphed. I've attempted multiplying and dividing the numerator and denominator by conjugates and of course taking the reciprocal of the natural log by removing the negative sign but I still can't figure out how to convert between one form and the other.

Any help would be appreciated, Thanks!
 
Eigendorf said:
Hello!

I'm trying to show that one function is equal to another but I'm not sure how to peform the operations to show that they are equal.

I'm fairly certain that the functions are equal because when I plug them both into my graphing calculator the results are the same.


This is part of a longer homework problem I'm working on and my answer is different from what is in the back of the book but the graphs of the function are the same.

Anyways the function that I'm trying to figure out how to convert to the answer in the back of the book is:

My Function:

\(\displaystyle -4\ln{\dfrac{4 + \sqrt{16 - x^2}}{x}}\)

The Books Function:

\(\displaystyle 4\ln{\dfrac{4 - \sqrt{16 - x^2}}{x}}\)


These appear to be equal when graphed. I've attempted multiplying and dividing the numerator and denominator by conjugates and of course taking the reciprocal of the natural log by removing the negative sign but I still can't figure out how to convert between one form and the other.

Any help would be appreciated, Thanks!
Click to expand...

What you describe sounds like just what I would do, and should work.

Please show the details -- what happened when you did that? You may have made some silly little error.
 
I figured out how to convert using trig functions but I don't understand an algebraic way to do it yet.

Basically if I convert the function

\(\displaystyle -4\ln{\dfrac{4 + \sqrt{16 - x^2}}{x}}\)

back into trig form


\(\displaystyle -4\ln{(csc{x} + cot{x})}\)

I can take the reciprocal

\(\displaystyle 4\ln{\dfrac{1}{csc{x} + cot{x}}}\)


Multiply by the conjugate

\(\displaystyle 4\ln{((\dfrac{1}{csc{x} + cot{x}}) * (\dfrac{csc{x} - cot{x}}{csc{x} - cot{x}}))}\)

Which becomes

\(\displaystyle 4\ln{\dfrac{csc{x} - cot{x}}{(csc{x})^2 - (cot{x})^2}}\)

And the Identity

\(\displaystyle {(csc{x})^2 - (cot{x})^2 = 1}\)

Which makes the function

\(\displaystyle 4\ln{(csc{x} - cot{x})}\)

and then I can convert it back to X values (the csc and cot should use theta but Idk how to show that in LaTex)

\(\displaystyle 4\ln{\dfrac{4 - \sqrt{16 - x^2}}{x}}\)


I don't know if there is a direct algebraic way to make that conversion but it seems that I can use trig functions to convert the form and then change back to X / Y values.
 
Eigendorf said:
I figured out how to convert using trig functions but I don't understand an algebraic way to do it yet.

Basically if I convert the function

\(\displaystyle -4\ln{\dfrac{4 + \sqrt{16 - x^2}}{x}}\)

back into trig form


\(\displaystyle -4\ln{(csc{x} + cot{x})}\)

I can take the reciprocal

\(\displaystyle 4\ln{\dfrac{1}{csc{x} + cot{x}}}\)


Multiply by the conjugate

\(\displaystyle 4\ln{((\dfrac{1}{csc{x} + cot{x}}) * (\dfrac{csc{x} - cot{x}}{csc{x} - cot{x}}))}\)

Which becomes

\(\displaystyle 4\ln{\dfrac{csc{x} - cot{x}}{(csc{x})^2 - (cot{x})^2}}\)

And the Identity

\(\displaystyle {(csc{x})^2 - (cot{x})^2 = 1}\)

Which makes the function

\(\displaystyle 4\ln{(csc{x} - cot{x})}\)

and then I can convert it back to X values (the csc and cot should use theta but Idk how to show that in LaTex)

\(\displaystyle 4\ln{\dfrac{4 - \sqrt{16 - x^2}}{x}}\)


I don't know if there is a direct algebraic way to make that conversion but it seems that I can use trig functions to convert the form and then change back to X / Y values.
Click to expand...

Do the exact same thing with the radical expression! Multiply numerator and denominator by the conjugate of the numerator, \(\displaystyle 4 - \sqrt{16 - x^2}\) . The numerator will simplify greatly and cancel with the denominator, leaving just what you want.
 
You do not need to use trig function to show they are the same.

Eigendorf said:
Hello!

I'm trying to show that one function is equal to another but I'm not sure how to peform the operations to show that they are equal.

I'm fairly certain that the functions are equal because when I plug them both into my graphing calculator the results are the same.


This is part of a longer homework problem I'm working on and my answer is different from what is in the back of the book but the graphs of the function are the same.

Anyways the function that I'm trying to figure out how to convert to the answer in the back of the book is:

My Function:

\(\displaystyle -4\ln{\dfrac{4 + \sqrt{16 - x^2}}{x}}\)

The Books Function:

\(\displaystyle 4\ln{\dfrac{4 - \sqrt{16 - x^2}}{x}}\)


These appear to be equal when graphed. I've attempted multiplying and dividing the numerator and denominator by conjugates and of course taking the reciprocal of the natural log by removing the negative sign but I still can't figure out how to convert between one form and the other.

Any help would be appreciated, Thanks!
Click to expand...


If you put - inside the ln, then you can use it to flip the numerator and denominator. If you rationalize the denominator, you will get the answer of the book.
 
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