# Derivation or proof of derivative sin (x)

• Mr X
In summary, you can use the Taylor series to derive a derivative for sin() and cos(), but it's not required.
Mr X
Homework Statement
Proof of sin(x) without using any other trig derivatives unproved.
I'm trying to derive the derivatives of trigonometric equations, while I can solve others using sine's derivative, I can't derive the derivative of sin x itself.
Relevant Equations
First principle, basic algebra and basic trig
How do I do this from here without using the derivatives of sin or cos ?

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It's not clear to me what you are allowed to use as "givens". But two things come to mind:

A point following a circle of radius 1 and centered at 0,0 will have it's x and y values change as functions of the sin and cos of time. That could be your starting point.

Or, if you are allowed to use the Taylor Series, then the game is over.

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Mr X
.Scott said:
It's clear to me what you are allowed to use as "givens". But two things come to mind:

A point following a circle of radius 1 and centered at 0,0 will have it's x and y values change as functions of the sin and cos of time. That could be your starting point.

Or, if you are allowed to use the Taylor Series, then the game is over.
Well I'm allowed to use anything, I'm just doing it for my own satisfaction. Tho i have no idea about Taylor series.is it possible to clear the basics in a few hours?
As for the other point while I've thought about it, I have no idea on how to apply the idea to the proof because I only have like mental picture and nothing more. especially when the /h come to play

Using a Taylor series is begging the question: you don't how that it has one until you know that it is diefferentiable.

The idea is that $$\frac{\sin(x + h)-\sin x}{h} = \frac{\sin x \cos h + \cos x \sin x}{h} = \sin x \frac{\cos(h) - 1}{h} + \cos x \frac{\sin h}{h}$$ and therefore $\sin$ is differentiable if $$\begin{split} \cos'(0) &= \lim_{h \to 0} \frac{\cos h - 1}{h}, \\ \sin'(0) &= \lim_{h \to 0} \frac{ \sin h}{h} \end{split}$$ both exist. You can estabish these by a geometric argument.

PhDeezNutz, Mr X, DaveE and 2 others
Using a Taylor series to derive a derivative is like using TNT to invent gunpowder.

Mr X, .Scott, DaveE and 1 other person
haushofer said:
Using a Taylor series to derive a derivative is like using TNT to invent gunpowder.
You could always define ##\sin## and ##\cos## using the Taylor expansion for ##e^{ix}## and Eulers formula.

Mr X
PeroK said:
You could always define ##\sin## and ##\cos## using the Taylor expansion for ##e^{ix}## and Eulers formula.

But then you'd have to prove that those functions satisfy the definitions in terms of ratios of sides of a right-angled triangle.

Mr X
It doesn't really make sense to "have to" prove a second definition, when you already have one. It may be enlightening, satisfying, etc, but it is not strictly necessary.

Mr X
Dragon27 said:
It doesn't really make sense to "have to" prove a second definition, when you already have one. It may be enlightening, satisfying, etc, but it is not strictly necessary.
It makes sense if you want to verify that your definitions using the Taylor series are the same as the precalculus trigonometry definitions of sin() and cos() that the OP was almost certainly referring to.

Mr X
Yes, it makes sense to relate back this new analytical definition to your previous geometrical intuition, which was not really rigorous (and developing rigorous geometry with angles and stuff so that you can define trigonometric function from the start is quite difficult, and it is much easier to work with analytical geometry anyway). I'm just not really buying this "you have to". It's more of a secondary consideration: "hey, you can calculate some geometrical stuff with these functions as well".

Mr X
Dragon27 said:
Yes, it makes sense to relate back this new analytical definition to your previous geometrical intuition, which was not really rigorous (and developing rigorous geometry with angles and stuff so that you can define trigonometric function from the start is quite difficult, and it is much easier to work with analytical geometry anyway). I'm just not really buying this "you have to". It's more of a secondary consideration: "hey, you can calculate some geometrical stuff with these functions as well".
To answer the OP question, it is best to deal with their question instead of a question that you wish had been asked.

Mr X and DaveE
FactChecker said:
To answer the OP question, it is best to deal with their question instead of a question that you wish had been asked.
@pasmith already answered the OP's question as it was likely intended, but it doesn't hurt to point out that a proof depends on how the trig functions are defined and that there is more than one way to define them.

Mr X and PeroK
Mr X said:
Homework Statement: Proof of sin(x) without using any other trig derivatives unproved.
I'm trying to derive the derivatives of trigonometric equations, while I can solve others using sine's derivative, I can't derive the derivative of sin x itself.
Relevant Equations: First principle, basic algebra and basic trig

How do I do this from here without using the derivatives of sin or cos ?
The limits that a left for your last step can be found without any derivatives.

Mr X
FactChecker said:
To answer the OP question, it is best to deal with their question instead of a question that you wish had been asked.
Well, tbh, I felt encouraged to try and probe more into the assumptions of the question by the fact that the assumptions were more guessed at, than clearly stated, and by this statement:
Mr X said:
Well I'm allowed to use anything, I'm just doing it for my own satisfaction.

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Mr X and FactChecker
Dragon27 said:
It doesn't really make sense to "have to" prove a second definition, when you already have one. It may be enlightening, satisfying, etc, but it is not strictly necessary.

One definition may be taken as axiomatically true; other equivalent definitions then become theorems which require proof.

Mr X and FactChecker
Another common relationship in an inner-product space is ##\vec{u}\cdot \vec{v} = \frac {\cos(\theta)}{|\vec{u}| |\vec{v}|}##. I have not seen that used to define ##\cos()##, but I have seen it used to define the angle ##\theta## between ##\vec u## and ##\vec v ## in an inner product space. I suppose that in those cases, the definition of ##\cos()## was analytical.

Mr X and vanhees71
I believe that if we cannot explain something to the students suitably because they have not got necessary general theory then we should wait until they will have it.
For example if we cannot prove ##\sin'=\cos## let the students accept it without proof and prove it when they will be ready.

We can define ##\cos## as a solution to the following IVP
$$\ddot x+x=0,\quad x(0)=1,\quad \dot x(0)=0;$$
and ##\sin## as a solution to the following IVP
$$\ddot y+y=0,\quad y(0)=0,\quad \dot y(0)=1.$$
These definitions give easily all the high-school trigonometry and all the analysis provided students attended a course of ODE previously.

vanhees71
Dragon27 said:
Yes, it makes sense to relate back this new analytical definition to your previous geometrical intuition, which was not really rigorous (and developing rigorous geometry with angles and stuff so that you can define trigonometric function from the start is quite difficult, and it is much easier to work with analytical geometry anyway). I
The projection of a unit vector onto the X-axis is easily made rigorous. I think it is just as "rigorous" and more intuitive than the power series. The power series requires a lot of work to make it "rigorous". To start with, limits and convergence need to be defined and established for that series.

Mr X and vanhees71
wrobel said:
I believe that if we cannot explain something to the students suitably because they have not got necessary general theory then we should wait until they will have it.
For example if we cannot prove ##\sin'=\cos## let the students accept it without proof and prove it when they will be ready.

We can define ##\cos## as a solution to the following IVP
$$\ddot x+x=0,\quad x(0)=1,\quad \dot x(0)=0;$$
and ##\sin## as a solution to the following IVP
$$\ddot y+y=0,\quad y(0)=0,\quad \dot y(0)=1.$$
These definitions give easily all the high-school trigonometry and all the analysis provided students attended a course of ODE previously.
Well, yes. For this you need some existence and uniqueness theory for ODEs, but that's not too difficult for these equations.

Mr X
FactChecker said:
The projection of a unit vector onto the X-axis is easily made rigorous. I think it is just as "rigorous" and more intuitive than the power series. The power series requires a lot of work to make it "rigorous". To start with, limits and convergence need to be defined and established for that series.
But you still have a lot of work to define angles rigorously. Analytical definition of sine/cosine is, imho, comparatively less effort.

Mr X and vanhees71
Dragon27 said:
But you still have a lot of work to define angles rigorously. Analytical definition of sine/cosine is, imho, comparatively less effort.
Not really. In an inner product space, the "angle" is the relative orientation of a unit vector that gives a certain inner product with a unit vector along the X-axis.

Mr X
FactChecker said:
In an inner product space, the "angle" is the relative orientation of a unit vector that gives a certain inner product with a unit vector along the X-axis.
How do you define angle measure?

Mr X
pasmith said:
But then you'd have to prove that those functions satisfy the definitions in terms of ratios of sides of a right-angled triangle.
That's the downside!

Mr X
DaveE said:
google search is a wonderful thing, sometimes better than PF.

Thank you, I understood that one.

and I have to say I cannot understand half of the replies in here, it really blew up. But well it is entertaining to read through them and many facts pointed out such as there are different definitions for the same functions of angle are interesting and kinda makes me wanna delve deeper. Tho I'll save looking up all that's here for another day after I've improved a little more. So thanks everyone

FactChecker and PeroK

## What is the derivative of sin(x)?

The derivative of sin(x) is cos(x).

## How do you derive the derivative of sin(x) using the limit definition?

To derive the derivative of sin(x) using the limit definition, we start with the limit definition of the derivative: $\frac{d}{dx} \sin(x) = \lim_{h \to 0} \frac{\sin(x+h) - \sin(x)}{h}$Using the sum-to-product identities, this becomes:$\frac{d}{dx} \sin(x) = \lim_{h \to 0} \frac{\sin(x)\cos(h) + \cos(x)\sin(h) - \sin(x)}{h}$$= \lim_{h \to 0} \left[ \sin(x) \frac{\cos(h) - 1}{h} + \cos(x) \frac{\sin(h)}{h} \right]$As $$h \to 0$$, $$\frac{\sin(h)}{h} \to 1$$ and $$\frac{\cos(h) - 1}{h} \to 0$$, so:$= \sin(x) \cdot 0 + \cos(x) \cdot 1$$= \cos(x)$Thus, the derivative of sin(x) is cos(x).

## Can you derive the derivative of sin(x) using the chain rule?

Yes, the chain rule can be used to derive the derivative of sin(x). Consider the function $$y = \sin(x)$$. Let $$u = x$$, then $$y = \sin(u)$$ and $$\frac{du}{dx} = 1$$. By the chain rule:$\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$Since $$\frac{dy}{du} = \cos(u)$$ and $$u = x$$, we get:$\frac{dy}{dx} = \cos(x) \cdot 1 = \cos(x)$Therefore, the derivative of sin(x) is cos(x).

## Why does the derivative of sin(x) equal cos(x)?

The derivative of sin(x) equals cos(x) because of the way the sine and cosine functions are defined and their properties in trigonometry. The derivative measures the rate of change of a function, and the rate of change of the sine function at any point x is given by the cosine of x. This relationship can be derived from the limit definition of the derivative or understood through the geometric interpretation of the unit circle, where the cosine

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