# Homework Help: Prove the differentiation rule: (cosx)' = sinx for any x.

1. Nov 15, 2011

### NWeid1

As the title says. I'm unsure what to use...limit definition? Thanks.

2. Nov 15, 2011

### Fredrik

Staff Emeritus

By the way, I don't like the notation (cos x)'. cos'(x) would be better.

3. Nov 15, 2011

### eumyang

Is that even true? The derivative of cosine is sine?

4. Nov 15, 2011

### Fredrik

Staff Emeritus
Good point. I didn't even notice that.

5. Nov 16, 2011

### Nano-Passion

He most likely meant that the derivative of cos x is -sin x.

OP, there is a couple way you can do this. You can do this using the one of the trig identities with chain rule but I'm not familiar with that route. A more familiar route is starting off with the definition of the derivative. As you go about doing the proof, try to manipulate the variables to fit a couple of the three special limits: cos and sin. Does that ring a bell?

The latter method is longer but I find it useful to figure out that way.

Last edited: Nov 16, 2011
6. Nov 16, 2011

### iRaid

Use the definition of a derivative, it should come out to -sinx. If you use the chain rule, you need to know what the derivative of cosine is so it's obviously not what he needs to do.

7. Nov 16, 2011

### Nano-Passion

No not cosine, you need to know what the derivative of sin is. After using a co-function identity where d/dx cos x = d\dx [sin pi/2 - x], you use the chain rule and manipulate accordingly.

8. Nov 16, 2011

### iRaid

y=cos(x)

y=cos(u) u=x
y'=-sinu u'=1

Therefore..
dy/dx=(-sinx)(1) = -sinx

9. Nov 16, 2011

### dextercioby

That's a circular argument. You're using in your proof what you should be proving...

10. Nov 16, 2011

### iRaid

That's what I'm saying, you cant use the chain rule if you need to prove cos'x = sinx, you should only use it when you need to solve equations like: y=cos3xtan4x

(not when proving the derivatives of trig functions)

11. Nov 16, 2011

write the definition of derivative, and use the fact that sinx~x for infinitely small values of x. you'll need Taylor series to prove that sinx~x but there is a nice way of proving that using binomial theorem that has been done by Euler. There is also a way of proving that approximation using geometry and high school algebra combined together.

Anyway, the key point is to use sinx~x when x approaches zero. once you know that, you will be able to prove that (cosx)'=-sinx and (sinx)'=cosx by writing down the formal definition of derivative. You also need to know that cos(x+y)=cosxcosy-sinxsiny and sin(x+y)=sinxcosy+sinycosx

Last edited: Nov 16, 2011
12. Nov 16, 2011

### lurflurf

by use of elementary trigonometry and continuity of sine it is obvious that

cos'(x)=-sin'(0) sin(x)=cos'(0)cos(x)-sin'(0)sin(x)

So sin'(0) needs to be found by some means, usually by use of the particular definition being used.

13. Nov 16, 2011

### HallsofIvy

Part of the problem is that how you would prove such a thing depends upon how you have defined "cos(x)". You cannot use the "trigonometry" definition- that requires that x be positive. One common definition is in terms of the unit circle: if you measure a distance t around the circumference of the unit circle, starting at the point (1, 0), the point you end at has, by definition, coordinates (cos(t), sin(t)). In that case, you can use trig identities to reduce $(cos(x+h)- cos(x))/h$ and limit properties. There are, however, problems without defining measurement around the circumference of a circle without using trig functions.

But you can also define cos(x) and sin(x) by the power series:
$$cos(x)= \sum_{n=0}^\infty \frac{(-1)^nx^{2n}}{(2n)!}$$
$$sin(x)= \sum_{n=0}^\infty \frac{(-1)^n x^{2n+1}}{(2n+1)!}$$
Because that power series converges for all x (which can be shown by the ratio test), it converges uniformly in every compact interval and so can be differentiated term by term:
$$cos'(x)= \sum_{n= 1}^\infty \frac{(-1)^n(2n)x^{2n-1}}{(2n)!}= \sum_{n=1}^\infty \frac{(-1)^nx^{2n-1}}{(2n-1)!}$$
Making the change of index, j= n-1, n=j+1 so $(-1)^n= (-1)^{j-1}= (-1)(-1)^j$, and 2n- 1= 2(j+1)- 1= 2j+ 1 so
$$cos'(x)=-\sum_{j=0}^\infty \frac{(-1)^jx^{2j+1}}{(2j+1)!}= -sin(x)$$

Another perfectly valid way to define sine and cosine is
"y= sin(x) is the function satisfying the differential equation y''= -y with intial conditions y(0)= 0, y'(0)= 1" and
"y= cos(x) is the function satisfying the differential equation y''= -y with initial conditions y(0)= 1, y'(0)= 0".

By the "existence and uniqueness theorem" for initial value problems we know that there exist unique functions satisfying those conditions. Further, since y''= -y is a second order linear differential equation, we know that any solution can be written as a linear combination of two independent solutions and it is easy to see that the functions above are independent solutions. That is, any solution to y''= -y can be written in the form y(x)= A cos(x)+ B sin(x).

In particular, if we write y(x)= cos'(x), then y'(x)= cos''(x)= -cos(x). Differentiating again, y''= -cos')(x)= -y. That is, this new function y satisfies that same differential equation, y''= -y and so y(x)= Acos(x)+ Bsin(x). But cos(x) satisfies the initial conditions cos(0)= 1, cos'(x)= 0 so y(0)= cos'(0)= 0 and y'(0)= -cos(0)= -1. Putting those into the formula for y, y(0)= Acos(0)+ Bsin(0)= A(1)+ B(0)= 0 so A= 0 while y'(0)= Acos'(0)+ Bsin'(0)= A(0)+ B(1)= -1 so B=-1. That is, y(x)= cos'(x)= 0(cos(x))+ (-1)(sin(x))= -sin(x).

Last edited by a moderator: Nov 16, 2011
14. Nov 16, 2011

### Fredrik

Staff Emeritus
Your argument was circular, but the method Nano-Passion suggested isn't. It uses the chain rule to find the derivative of cos, but to use it, you must know the derivative of sin.

This method is slightly more complicated than I first thought. To use it, you must be able to find the limits of $(\cos h-1)/h$ and $(\sin h)/h$ as $h\to 0$.

It took me a pretty long time to figure out what you're doing here. I still don't see how you got that expression in the middle (unless you got it by using cos'(0)=0 on the right-hand side). So I think the OP will find this pretty difficult. But it is a nice method.

15. Nov 16, 2011

Sure. That's why I told him that he needs to know sinx~x when x approaches 0. that's the same as saying sinh/h~1 for infinitely small values of h. also, (cosh-1)/h can be found once you know sinx~x.

16. Nov 16, 2011

### lurflurf

The difference identity

$$\cos(A)-\cos(B)=-2 \sin\left( \frac{A}{2}-\frac{B}{2}\right) \sin\left( \frac{A}{2}+\frac{B}{2}\right)$$

by use of the difference identity of by manipulation of the addition identity we have

$$\frac{ \cos(x+h)-\cos(h) }{h}= -\frac{ \sin\left( \frac{h}{2}\right)}{\frac{h}{2}} \sin\left( x+\frac{h}{2}\right)\rightarrow - \sin'(0) \sin(x)$$

by direct application of the addition identity

$$\frac{ \cos(x+h)-\cos(h) }{h}= \left( \frac{\cos(h)-1}{h} \right) \cos(x)-\left( \frac{\sin(h)}{h}\right) \sin(x) \rightarrow cos'(0)cos(x)-\sin(0) \sin'(x)$$

so comparison of the two not only establishes cos'(0)=0, but also the fact that cos'(0) is in a sense obvious in a way sin'(0) is not.

17. Nov 16, 2011

### Nano-Passion

Its pretty simple if you memorized it. In my calculus book they were introduced as "three special limits" and were required to be memorized, where a couple included$(\cos h-1)/h$ and $(\sin h)/h$ as $h\to 0$ the third one was $\lim_{x\ to\ 0} (1 + x) ^{1/x} = e$; of which was used in proving the derivative of e^x.

$$\lim_{x\to 0} \frac{cos x - 1}{x} = 0$$
$$\lim_{x\to 0} \frac{sin x}{x} = 1$$

Using the chain rule is the easiest method. But OP might feel more comfortable learning it by the limit process because that is what would ring a bell if this problem was on a test.

18. Nov 16, 2011

### Nano-Passion

Huh? They are used in the proof by the limit process and proof by the chain rule? What you really mean is that it isn't a rigorous proof and wouldn't be a proper one correct?

19. Nov 16, 2011

### Fredrik

Staff Emeritus
Ah, then I wasn't even close when I tried to guess what you were doing. (I actually don't even recognize that identity). I thought you were just taking derivatives of cleverly chosen trigonometric identities and solving for cos'(x). For example like this:

cos(x+y)=cos x cos y-sin x sin y.

Note that

d/dx cos(x+y) = cos'(x+y) =d/dy cos(x+y).

This implies that

cos'(x) cos y-sin'(x) sin y = cos x cos'(y)-sin x sin'(y).

Now choose y=0.

cos'(x) = cos x cos'(0) - sin x sin'(0)

20. Nov 16, 2011

### Nano-Passion

Do you mean $$cos'(0)cos(x)-\sin'(0) \sin(x)$$

Here is my portion of the proof in case it is more comforting.

We will use the sum & difference rule where $$cos(u±v) = cosu cosv \mp sinu sin v$$

$$\frac{d}{dx} [cos x] = \frac{cos (x+\delta x) - cos x}{\delta x}$$
$$= \frac{cos x cos \delta x - sin x sin \delta x - cos x}{\delta x}$$
$$=\frac{ -sin x sin \delta x +cos x cos \delta x - cos x}{\delta x}$$
$$=\frac{-sin x sin \delta x + cos x (1- cos \delta x)}{\delta x}$$
$$= - sin x (\frac{sin \delta x}{x}) + cos x (\frac{1 - cos \delta x}{\delta x})$$
$$= - sin x \lim_{\delta x\to 0}(\frac{sin \delta x}{x}) + cos x \lim_{\delta x\to 0} (\frac {1 - cos \delta x}{\delta x}$$
$$= - sin (1) - cos (0)$$
$$= - sin x$$

21. Nov 16, 2011

### Fredrik

Staff Emeritus
Nano, you need to type \sin and \cos instead of sin and cos when you use LaTeX. A LaTeX guide for the forums can be found here if you need it. (Hm, maybe it should say that you need to type \sin and \cos right at the beginning).

Oh yeah, one more thing: We don't normally post complete solutions when someone asks for help with a textbook-style problem. But since too much information had already been revealed, I don't think it matters here.

Edit: You can edit your post to insert the \ symbols until 11 hours and 40 minutes (700 minutes) after you posted.

Last edited: Nov 16, 2011
22. Nov 16, 2011

### Nano-Passion

Oh okay thank you.

Test 1
$$\cos x$$

23. Nov 17, 2011

### HallsofIvy

You seem to be referring to several different points here. When I said that "you cannot use the trigonometry definition- that requires that x be positive" I was referring to the definiton of sine as "opposite side over hypotenuse" and cosine as "near side over hypotenuse" in a right triangle. In those definitions, the variable is an angle and must be between 0 and $\pi/2$ radians.

24. Nov 17, 2011

but if you interpret sine and cosine functions as orthogonal coordinates that we're projecting the circle x^2 + y^2 = 1 on them then there will be no problems interpreting negative angles geometrically. you just need to reverse the direction from counter-clockwise to clockwise.

It's very easy to prove this equation. all you need to know is that sin(x)~x when x is infinitely small. well, a simple intuitive proof for this can be to graph y= sin(x) and the line y=x. when x is infinitely small, values of these two functions are becoming closer and closer to each other because y=x is tanget to y=sin(x) at x=0. Hence, $lim_{x→0}\sin{x}=lim_{x→0}x$. so we can say $lim_{x→0} \frac{\sin{x}}{x}=1$.
From trig. identities (sort of haversine formula) we have: $2sin^2(\frac{x}{2})=1-cos(x)$, hence, if we take limits from both sides and let x approach zero, we'll have:
$$lim_{x→ 0} {(2 sin(\frac{x}{2})sin(\frac{x}{2}))} = lim_{x→0}{(1-cos(x))}$$
$$lim_{x→ 0} {(2 \frac{x}{2} \frac{x}{2})} = lim_{x→0}{(1-cos(x))}$$
$$lim_{x→ 0} {\frac{(1-cos(x))}{x}}= lim_{x→ 0}{\frac{(2 \frac{x}{2} \frac{x}{2})}{x}}$$
$$lim_{x→ 0}{\frac{2 \frac{x}{2} \frac{x}{2}}{x}}=lim_{x→ 0}{x/2}=0$$
Hence,
$$lim_{x→ 0}{\frac{1-cos(x)}{x}}=0$$

Knowing these two limits and the trig. identities sin(x+y)=sinxcosy+sinycosx cos(x+y)=cosxcosy-sinxsiny, you can prove (cosx)'=-sinx and (sinx)'=cosx, just like the way Nano-Passion proved it, And once you know the derivatives of sine and cosine, you can easily find the derivatives of all trigonometric functions (tan,cot,sec,csc,...)

Last edited: Nov 17, 2011
25. Nov 17, 2011

### Fredrik

Staff Emeritus
AdrianZ, the result will look better if you type \lim, \sin and \cos instead of lim, sin and cos.