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NWeid1
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As the title says. I'm unsure what to use...limit definition? Thanks.
Is that even true? The derivative of cosine is sine?Prove the differentiation rule: (cosx)' = sinx for any x.
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.
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.
y=cos(x)
y=cos(u) u=x
y'=-sinu u'=1
Therefore..
dy/dx=(-sinx)(1) = -sinx
That's a circular argument. You're using in your proof what you should be proving...
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.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=cos^{3}xtan4x
(not when proving the derivatives of trig functions)
This method is slightly more complicated than I first thought. To use it, you must be able to find the limits of [itex](\cos h-1)/h[/itex] and [itex](\sin h)/h[/itex] as [itex]h\to 0[/itex].write the definition of derivative, and use the fact that sinx~x for infinitely small values of x.
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.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.
This method is slightly more complicated than I first thought. To use it, you must be able to find the limits of [itex](\cos h-1)/h[/itex] and [itex](\sin h)/h[/itex] as [itex]h\to 0[/itex].
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).
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 [itex](\cos h-1)/h[/itex] and [itex](\sin h)/h[/itex] as [itex]h\to 0[/itex].
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.
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 [itex](cos(x+h)- cos(x))/h[/itex] and limit properties.
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:The difference identity
...
The difference identity
[tex]\cos(A)-\cos(B)=-2 \sin\left( \frac{A}{2}-\frac{B}{2}\right) \sin\left( \frac{A}{2}+\frac{B}{2}\right)[/tex]
by use of the difference identity of by manipulation of the addition identity we have
[tex]\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)[/tex]
by direct application of the addition identity
[tex]\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)[/tex]
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.
Oh okay thank you.Nano, you need to type \sin and \cos instead of sin and cos when you use latex.
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 [itex]\pi/2[/itex] radians.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?
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 [itex]\pi/2[/itex] radians.
AdrianZ, the result will look better if you type \lim, \sin and \cos instead of lim, sin and cos.
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 [itex]\pi/2[/itex] radians.
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.
Oh, so you are referring to the definition sine and cosine on the unit circle. I was referring to not having "negative" angle in a right triangle. The difficulty with the unit circle definition is that it requires the circumference around the circle. And that requires an integral that inevitably leads to use of sine and cosine! Indeed that integral requires integrating [itex]1/\sqrt{1- x^2}[/itex] which is undefined at that starting point, x= 1, y= 0. It is an improper integral and just proving the integral exists require some use of sine and cosine. So trying to define sine and cosine in that way is definitely circular!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.
I don't think that's accurate. See disregardthat's post #79 in the thread where we discussed these things in August: Link.Indeed that integral requires integrating [itex]1/\sqrt{1- x^2}[/itex] which is undefined at that starting point, x= 1, y= 0. It is an improper integral and just proving the integral exists require some use of sine and cosine. So trying to define sine and cosine in that way is definitely circular!