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In my old age I decided to take up calc for a hobby. I need some expertise on a trig/calc issue.

Im working on proofs for the derivatives of the sine and cosine functions and having a bit of difficulty understanding one part of it.

So, as you know, using the regular difference quotient formula (aka the derivative), you would have:

f' sin x = (sin (x + h) - sin x)/ h

and we know that the formula for the sine of the sum of two angles is sin A cos B +cos A sin b.

So you can substitute that into the difference quotient and get
f' sin x = lim as h approaches 0 of
(sin x cos h + cos x sin h - sin x)/h

now here is what I don't get. you can separate the terms in the numerator, and come up with the following:

dy/dx = lim x as h approaches 0
sin x (cos h - 1/h) + lim x as h approaches 0 cos x sin h/h.

My problem is that I cannot for the life of me figure out where that -1 is coming from.

Do you know?
 
Posts: 283 | Registered: 02 November 2012Reply With Quote
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Well, I'm really good in basic arithmetic, but I forgot all the algebra and calculus I ever wanted to know, which was none, about 45 years ago.

So I emailed your post to my son, who has never gotten less than an A in any math course and is breezing thru college calculus and he sent me this answer. Whether it's right or not, or whether you'll be able to understand it, is another matter, because I sure as hell don't.

quote:
Well, you start with the equation:

f' sin x= limit as h approaches 0 of (sin x cos h + cos x sin h - sinx)/ h

This can be split up into:

f' sin x= limit as h approaches 0 of (sin x cos h - sin x)/h + limit as h approaches 0 of (cos x sin h)/h

Factoring out the sin x from the first half gives you:

f' sin x= limit as h approaches 0 of (sin x) (cos h -1)/h + limit as h approaches 0 of (cos x sin h)/h

The (cos h - 1) comes from dividing sin x out of (sin x cos h - sin x) because (sin x)/(sin x)=1.


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Posts: 17099 | Location: Texas USA | Registered: 07 May 2001Reply With Quote
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Basically, what it is, is looking at the instantaneous rate of change (derivative) of the Sine function (the ratio of the opposite side of an angle theta (often called sin x) to the hypoteneuse in a right triangle). (Answer turns out to be Cosine X).

And yes, I should have seen that.

Thank you very much. I understand it now. Tell him thank you.
 
Posts: 283 | Registered: 02 November 2012Reply With Quote
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You dropped a sign with you factored out the sin.
 
Posts: 3034 | Location: Colorado | Registered: 01 July 2010Reply With Quote
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