What is the missing function in this trigonometric proof?

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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!