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mf2004
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Hello

So if we take sinx and try to differnentiate from first principle :

dy/dx =lim (sin(x+h)-sinx)/h
dy/dx =lim (sinxcosh+sinhcosx-sinx)/h

cosh=1

=>

dy/dx = lim sinhcosx/h

now we have to use lim h-->0 sinh/h =1, but how can we know this limit if we can't do a Taylor series (we don't know what dsinx/dx is yet so we can't expand), l'Hôpital's rule won't work either :confused:

I'm confused as to how you would go around proving it so if anyone has any suggestions.

Thanks
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SimonM
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You might need to know \displaystyle \lim_{h \to 0} \frac{\sin h}{h} = 1
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ukdragon37
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(Original post by mf2004)
Hello

So if we take sinx and try to differnentiate from first principle :

dy/dx =lim (sin(x+h)-sinx)/h
dy/dx =lim (sinxcosh+sinhcosx-sinx)/h

cosh=1

=>

dy/dx = lim sinhcosx/h

now we have to use lim h-->0 sinh/h =1, but how can we know this limit if we can't do a Taylor series (we don't know what dsinx/dx is yet so we can't expand), l'Hôpital's rule won't work either :confused:

I'm confused as to how you would go around proving it so if anyone has any suggestions.

Thanks
Draw a unit quarter circle and use the squeeze theorem. I'll post a proof in a minute.
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Punk Phloyd
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(Original post by mf2004)
Hello

So if we take sinx and try to differnentiate from first principle :

dy/dx =lim (sin(x+h)-sinx)/h
dy/dx =lim (sinxcosh+sinhcosx-sinx)/h

cosh=1

=>

dy/dx = lim sinhcosx/h

now we have to use lim h-->0 sinh/h =1, but how can we know this limit if we can't do a Taylor series (we don't know what dsinx/dx is yet so we can't expand), l'Hôpital's rule won't work either :confused:

I'm confused as to how you would go around proving it so if anyone has any suggestions.

Thanks
Why not?
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bballer4life
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differentiate the series version of sin x?
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ukdragon37
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(Original post by Punk Phloyd)
Why not?
(Original post by bballer4life)
differentiate the series version of sin x?
They depend on differentiation of sin x, which is what you are proving so the argument is circular.

You must use an alternate method not employing differentiation.
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jhwilliam
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I think it can be proved by use of Exponentials and complex numbers although im not too familiar with the proof myself.
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bballer4life
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(Original post by ukdragon37)
They depend on differentiation of sin x, which is what you are proving so the argument is circular.

You must use an alternate method not employing differentiation.


sinx can be written as a power series in terms of dofferent powers of x, just differentiate it, and you get the power series version on cos x

whats wrong with that. i'm not making any a ssumptions
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ukdragon37
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(Original post by bballer4life)
sinx can be written as a power series in terms of dofferent powers of x, just differentiate it, and you get the power series version on cos x

whats wrong with that. i'm not making any a ssumptions
And how do you justify the power series? By differentiation and finding co-efficients of each term. Think of it this way. If you don't know what the integral or derivative of sinx is, how would you find the power series?
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DFranklin
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(Original post by bballer4life)
sinx can be written as a power series in terms of dofferent powers of x, just differentiate it, and you get the power series version on cos x

whats wrong with that. i'm not making any a ssumptions
Well, you made either an assumption or a definition when you said:

"sinx can be written as a power series in terms of dofferent powers of x"

(A common approach at university level is to use the power series as a definition, which is fine, but you then really need to prove that the function so defined behaves like the "old fashioned" version of sin. Which most universities don't bother to do).
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paronomase
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When h tends to 0, cosh-1 tends to 0, and sinh tends to 0
You have your answer.
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paronomase
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oups, sorry, I forgot to mention that sinh/h = 1 as h tends to 0
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ukdragon37
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(Original post by paronomase)
oups, sorry, I forgot to mention that sinh/h = 1 as h tends to 0
and how do you prove that?
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ukdragon37
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(Original post by mf2004)
...


This is a quarter circle with radius 1. Call the angle of the smaller sector x. By the properties of the unit circle we observe that the red triangle has area 0.5sinx. The red+green sector has area x/(2pi)*pi=x/2. The yellow+green+red triangle has area 0.5tanx. (why?)

The area of the sector is between that of the two triangles, so we know that, assuming x is not 0:

\begin{array}{l}

 \frac{1}{2}\sin x < \frac{x}{2} < \frac{1}{2}\tan x \\ 

 \sin x < x < \tan x \\ 

 1 < \frac{x}{{\sin x}} < \frac{1}{{\cos x}} \\ 

 \end{array}

By inverting we have:

\begin{array}{l}

 \frac{{\sin x}}{x} < 1 \\ 

 \frac{{\sin x}}{x} > \cos x \\ 

 \cos x < \frac{{\sin x}}{x} < 1 \\ 

 \mathop {\lim }\limits_{x \to 0} \cos x < \mathop {\lim }\limits_{x \to 0} \frac{{\sin x}}{x} < 1 \\ 

 \mathop {\lim }\limits_{x \to 0} \cos x = 1 \\ 

 \end{array}


Since the limit of sinx/x is between that of cos x and 1 and they both tend to 1, by the squeeze theorem sinx/x must tend to 1 also.
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bballer4life
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(Original post by ukdragon37)
And how do you justify the power series? By differentiation and finding co-efficients of each term. Think of it this way. If you don't know what the integral or derivative of sinx is, how would you find the power series?

ah, i see what you mean. since you have to use the taylor expansion to obtain the power series, it can't be used to prove the differential sinx

yeah i guess using the formal definition is the best way
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paronomase
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lim sinx/x = sinx - sin 0 / x - 0 = sin'(0) = cos (0) = 1
I agree that you cannot use this method, since you dont know that sin'(x) = cos (x)

try this one then
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paronomase
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But it is a result you normally prove once in class, and then take for granted. You do not have to prove it again every time you have a trig exercise
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ukdragon37
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(Original post by paronomase)
But it is a result you normally prove once in class, and then take for granted. You do not have to prove it again every time you have a trig exercise
The title of this thread is actually ill-posed. In the original post OP actually wanted a proof of the limit of sinx/x.
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DFranklin
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Why do you say sin x < x < tan x? (It's worth pointing out that many "proofs" of these inequalities are dubious, from a rigourous point of view. For example, anything relying on arclengths is questionable, since it is very hard to give a mathematical definition of arclength that doesn't rely on a lot of other machinery).
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ukdragon37
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(Original post by DFranklin)
Why do you say sin x < x < tan x? (It's worth pointing out that many "proofs" of these inequalities are dubious, from a rigourous point of view. For example, anything relying on arclengths is questionable, since it is very hard to give a mathematical definition of arclength that doesn't rely on a lot of other machinery).
Me?

Well, I'm only saying it's valid within the region of the quarter circle.
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