What is the missing function in this trigonometric proof?

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Homework Help Overview

The discussion revolves around a trigonometric proof involving the expression (cot5x * [trigonometric function]5x)/csc6x, which the original poster attempts to prove equals 5/6. The missing function in the expression is a point of focus.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants explore various trigonometric functions to identify the missing component, with one suggesting sin5x and another proposing sec(5x) based on limit evaluations. Questions arise regarding the correctness of the initial expression and its equivalences.

Discussion Status

Some participants provide insights into the structure of the expression and its transformations, while others clarify the limits involved. The original poster expresses uncertainty about the missing function, and there is a shift in understanding regarding the problem's requirements.

Contextual Notes

There is mention of a potential misstatement in the problem, indicating that the goal may have been to prove the function equals 6/5 instead of 5/6, which adds complexity to the discussion.

matthewd49
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Homework Statement



basically i have this proof to do where (cot5x * [trigonometric function]5x)/csc6x = 5/6 and i have to prove it equals 5/6

Homework Equations



but what i can't remember about the problem is the second function. I've tried plugging in all of them and using a made up value of x = 10 just to see if i could figure out what the function i can't remember it is but i haven't had any luck.

cos(0) = 1
sin(x)/x = 1

The Attempt at a Solution



the closest I've come to what may be the missing function is when i plug in sin5x but i get a number that is approximately .85 and not .833.

well if i could figure out the function i know to set it up like this ((cos5x/tan5x) * [unknown function]5x)/(1/sin6x)
 
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are you evaulating a limit as x tends to zero?

in general, for arbitrary [itex]x \neq 2 n\pi[/itex]
[tex]\frac{sin(x)}{x}\neq 1[/tex]

And in fact when [itex]x =n\pi[/itex] the function is undefined, but the limit exists, for example
[tex]\lim_{x \to 0}\frac{sin(x)}{x}= 1[/tex]
 
matthewd49 said:

Homework Statement



basically i have this proof to do where (cot5x * [trigonometric function]5x)/csc6x = 5/6 and i have to prove it equals 5/6
...

the closest I've come to what may be the missing function is when i plug in sin5x but i get a number that is approximately .85 and not .833.

well if i could figure out the function i know to set it up like this ((cos5x/tan5x) * [unknown function]5x)/(1/sin6x)
Hello matthewd49. Welcome to PF !

How sure are you that the expression was of the form [itex]\displaystyle\frac{\cot(5x)\text{trig}(5x)}{\csc(6x)}\,,[/itex] where trig(θ) is one of the trig functions?

This expression is equivalent to [itex]\displaystyle\frac{\cos(5x)\text{trig}(5x)\sin(6x)}{\sin(5x)}\,,[/itex] also [itex]\displaystyle\frac{\text{trig}(5x)\sin(6x)}{\tan(5x)}\,.[/itex]

You also have [itex]\frac{\displaystyle\frac{\cos(5x)}{ \tan(5x)}\text{trig}(5x)}{\displaystyle\frac{1}{ \sin(6x)}}\,,[/itex] which is not equivalent to the above expression.

Your last expression is equivalent to [itex]\displaystyle\frac{\cos^2(5x)\text{trig}(5x)\sin(6x)}{\sin(5x)}\,.[/itex]

BTW: [itex]\displaystyle\lim_{x\,\to\,0}\frac{\sin(6x)}{\sin(5x)}=\frac{6}{5}\,.[/itex] This leads me to believe that your mystery function is [itex]\sec(5x)\,.[/itex]
 
hi guys thanks for all your help. i found out the missing function was actually sec5x and then found out afterwards that they had presented the problem wrong and wanted me to prove that said function was = to 6/5, not 5/6. that made it a much easier problem which i quickly finished. thanks for all of your help though, you guys are awesome!
 

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