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# Differentiating sin x : sin δx ≈ δx Tweet

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1. Differentiating sin x : sin δx ≈ δx
[Solved]

I'm just looking at the topic mentioned in the title. I don't fully understand it. I know that the gradient of the sine curve at the origin is 1 which means that the tangent at that point has the equation y = x. Which shows how sin δx ≈ δx. Now what I would like to ask for is a mathematical explanation of why the gradient of sin x at x=0 is 1. I know that this is true and that differentiating sin x gives cos x and cos(0)=1 but could someone provide an explanation other than this?
Last edited by Ateo; 01-08-2012 at 17:37.
2. Re: Differentiating sin x : sin δx ≈ δx
The gradient of a curve is 'the best straight line approximation to the curve at that point'.

To find this, we draw a straight line between sin(0) and sin(x) for some fixed value of x close to 0 (say 0.1). The gradient of this line is - this just comes from the standard equation for the gradient of a straight line.

Now we want to let x get closer and closer to 0. This will give a straight line on the graph which gets closer and closer to the tangent to the graph at x=0. Mathematically, this is the same as . It can be proved that in fact, as x gets close to 0, gets close to 1. So that's our answer - the gradient is 1.

This is done algebraically to prove that the derivative of sin is cos, and the same procedure is the method for finding the derivatives of all the other functions that you know. Is that what you wanted to know?
3. Re: Differentiating sin x : sin δx ≈ δx
(Original post by Ateo)
I'm just looking at the topic mentioned in the title. I don't fully understand it. I know that the gradient of the sine curve at the origin is 1 which means that the tangent at that point has the equation y = x. Which shows how sin δx ≈ δx. Now what I would like to ask for is a mathematical explanation of why the gradient of sin x at x=0 is 1. I know that this is true and that differentiating sin x gives cos x and cos(0)=1 but could someone provide an explanation other than this?
The way I was introduced to the analysis of trig functions came from the other direction. In this construction, we define

and

then we show certain convergence properties are satisfied so that we can swap and , so that

This works, but the construction we used requires you to prove that what we define sin and cos to be and what we want sin and cos to be are the same thing... which is another matter in itself!

Edit: I'm not sure whether I've really answered your question. This explains how we know that the derivative of sine is cosine, of which the derivative of sin x at 0 being 1 is a consequence; and it's a necessary result for applying l'Hôpital's rule, as in the post below mine... but maybe you were allowed to assume this. Shrug.
Last edited by nuodai; 01-08-2012 at 17:56.
4. Re: Differentiating sin x : sin δx ≈ δx
As said above, by definition the gradient is:

To show that this is equal to 1, you use something called l'Hôpital's rule (http://en.wikipedia.org/wiki/L'H%C3%B4pital's_rule) which tells you:

5. Re: Differentiating sin x : sin δx ≈ δx
(Original post by nuodai)
The way I was introduced to the analysis of trig functions came from the other direction. In this construction, we define

and

Thanks for this alternative method. Didn't you mean this though:

and

6. Re: Differentiating sin x : sin δx ≈ δx
(Original post by Ateo)
I'm just looking at the topic mentioned in the title. I don't fully understand it. I know that the gradient of the sine curve at the origin is 1 which means that the tangent at that point has the equation y = x. Which shows how sin δx ≈ δx. Now what I would like to ask for is a mathematical explanation of why the gradient of sin x at x=0 is 1. I know that this is true and that differentiating sin x gives cos x and cos(0)=1 but could someone provide an explanation other than this?
What do you want an explanation of? Why d/dx(sin x) = cos x ? Or why we are interested in why sin x is like x when x is small ? Or both ?

You will probably find it easiest to consult a decent textbook that describes "differentiation from first principles", and can give you some decent diagrams, if you've never come across this kind of thing before. Alternatively a google search will turn up all kinds of resources that explain this; the Khan Academy has plenty of videos that deal with elementary differentiation and limits, for example.

In addition, if you're specifically interested in the derivative of sin x, you need to be sure that you're comfortable with the idea of radians, and that you can sketch out (for example) sin x when x is measured in both radians and degrees. That' s because d/dx(sin x) = cos x, only if x is measured in radians (if x is measured in degrees, you need a scale factor of pi/180 to make it true - this becomes obvious if you sketch out little slope triangles on a sin curve, with the x axis measured in radians, then in degrees)
7. Re: Differentiating sin x : sin δx ≈ δx
(Original post by dantheman1261;38782419 It can be proved that in fact, as x gets close to 0, [latex)
\frac{sin(x)}{x} [/latex] gets close to 1. So that's our answer - the gradient is 1.
It is that proof that I was looking for, thanks I understand it now.
8. Re: Differentiating sin x : sin δx ≈ δx
(Original post by nuodai)
The way I was introduced to the analysis of trig functions came from the other direction. In this construction, we define

and

And how would you construct those series, without first knowing how to differentiate sin and cos?
9. Re: Differentiating sin x : sin δx ≈ δx
(Original post by atsruser)
What do you want an explanation of? Why d/dx(sin x) = cos x ? Or why we are interested in why sin x is like x when x is small ? Or both ?

You will probably find it easiest to consult a decent textbook that describes "differentiation from first principles", and can give you some decent diagrams, if you've never come across this kind of thing before. Alternatively a google search will turn up all kinds of resources that explain this; the Khan Academy has plenty of videos that deal with elementary differentiation and limits, for example.

In addition, if you're specifically interested in the derivative of sin x, you need to be sure that you're comfortable with the idea of radians, and that you can sketch out (for example) sin x when x is measured in both radians and degrees. That' s because d/dx(sin x) = cos x, only if x is measured in radians (if x is measured in degrees, you need a scale factor of pi/180 to make it true - this becomes obvious if you sketch out little slope triangles on a sin curve, with the x axis measured in radians, then in degrees)
This is what I was looking for:

10. Re: Differentiating sin x : sin δx ≈ δx
(for x small and positive)

Now divide through by sin x:

Then as x tends to 0, 1/cos x tends to 1. Thus, by the squeeze rule, x/sin x tends to 1. So (sin x)/x tends to 1.
11. Re: Differentiating sin x : sin δx ≈ δx
(Original post by Lord of the Flies)
As said above, by definition the gradient is:
To show that this is equal to 1, you use something called l'Hôpital's rule
l'Hôpital's rule involves already knowing how to differentiate sine - and knowing how to differentiate sine involves using the limit you're trying to show
12. Re: Differentiating sin x : sin δx ≈ δx
(Original post by Ateo)
Thanks for this alternative method. Didn't you mean this though:
Woops! Yes I did.
(Original post by atsruser)
And how would you construct those series, without first knowing how to differentiate sin and cos?
It comes from properties of the exponential function and complex numbers. This result was almost certainly known before these constructions were made, but then to make everything rigorous we 'worked backwards' and then showed that what we did was equivalent. (Much of mathematical analysis involves taking what you believe, turning it around and showing from mathematically-basic-but-conceptually-tricky principles that what you believe is consistent.)
13. Re: Differentiating sin x : sin δx ≈ δx
(Original post by nuodai)
Woops! Yes I did.

It comes from properties of the exponential function and complex numbers. This result was almost certainly known before these constructions were made, but then to make everything rigorous we 'worked backwards' and then showed that what we did was equivalent. (Much of mathematical analysis involves taking what you believe, turning it around and showing from mathematically-basic-but-conceptually-tricky principles that what you believe is consistent.)
Yes, I guess strictly you are right: there is indeed a calculus-free argument that constructs two functions and from the real and complex parts of the series expansion of , and where and can be shown to have the required properties to be cos and sin, but that arguments also requires:

- knowledge of the Taylor series of and the fact that it converges
- the assumption that we can construct a valid complex power series from it
- the assumption that we can differentiate the real and complex parts of the series term-by-term to produce new convergent series
- a subtle series of arguments that show those power series do indeed behave like cos and sin

I think that's a little much for someone who is still learning how to differentiate elementary functions, and without all of that derivation, it's fairly pointless to present the power series as fait-accomplis. They're not really comprehensible without a fair bit of mathematical sophistication, I think.