My Math Forum How to prove this question?

 Calculus Calculus Math Forum

 December 21st, 2013, 06:27 PM #1 Senior Member   Joined: Sep 2013 From: Earth Posts: 827 Thanks: 36 How to prove this question? Prove that d(sin(x))/dx=cos(x),
 December 21st, 2013, 07:05 PM #2 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,898 Thanks: 1093 Math Focus: Elementary mathematics and beyond Re: How to prove this question? $\lim_{h\to0}\,\frac{\sin(x\,+\,h)\,-\,\sin(x)}{h}$ $=\,\lim_{h\to0}\,\frac{\sin(x)\cos(h)\,+\,\sin(h)\ cos(x)\,-\,\sin(x)}{h}$ $=\,\sin(x)\lim_{h\to0}\,\frac{\cos(h)\,-\,1}{h}\,+\,\cos(x)\lim_{h\to0}\,\frac{\sin(h)}{h}$ $=\,\sin(x)\,\cdot\,0\,+\,\cos(x)\,\cdot\,1\,=\,\co s(x)$
 December 21st, 2013, 08:03 PM #3 Math Team   Joined: Sep 2007 Posts: 2,409 Thanks: 6 Re: How to prove this question? Using what definition of "sin(x)" and cos(x)? greg1313 showed how to do it if you are given some definiton, such as the definition in terms of unit circles. that lead immediately to sin(a+ b)= sin(a)cos(b)cos(a)sin(b) and that sine and cosine are continuous at 0. But there are logical difficulties with defining "arc length" for a unit circle before you have defined sine and cosine so some texts use "y= sin(x) is the function satisfying y''+ y= 0 with initial conditions y(0)= 0, y'(0)= 1 and y= cos(x) is the function satisfying y''+ y= 0 with initial conditions y(0)= 1, y'(0)= 0. Using that definition, if y= (sin(x))', then y'= (sin(x))''= -sin(x) (because (sin(x))''+ sin(x)= 0) so y''= - sin'(x)= -y so that y''+ y= 0. Further y(0)= sin'(0)= 1 and sin''(0)= -y(0)= 0. Since y= (sin(x))' satisfies y''+ y= 0, y(0)= 1, y'(0)= 0, y'= (sin(x))'= cos(x). Yet another perfectly valid definition of sin(x) and cos(x) is $sin(x)= \sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!}x^{2n+1}$ and $cos(x)= \sum_{n=0}^\infty \frac{(-1)^n}{(2n)!}x^{2n}$ We show that sin(x) converges for all x so converges uniformly in any closed and bounded interval. That means that, in particular, sin(x) is "term by term differentiable" at any x. So $(sin(x))'= \sum_{n=0}^\infty \frac{1}{(2n+1)!}(2n+ 1)x^{2n}= \sum_{n=0}^\infty \frac{1}{(2n)!}x^{2n}= cos(x)$
December 22nd, 2013, 09:49 AM   #4
Global Moderator

Joined: Oct 2008
From: London, Ontario, Canada - The Forest City

Posts: 7,898
Thanks: 1093

Math Focus: Elementary mathematics and beyond
Re: How to prove this question?

Quote:
 Originally Posted by HallsofIvy But there are logical difficulties with defining "arc length" for a unit circle before you have defined sine and cosine . . .
Such as? If we let one radian be the angle subtending the arc length equivalent to the radius of the unit circle, then arc length = r? with r being the radius of the circle and ? being the angle subtending the arc. A geometric interpretation of the limit sin(?)/? can now be given: as ? approaches 0, sin(?) approaches the arc length of ? radians. This can be visualized as the arc length getting closer and closer to a straight line, that is, closer and closer to sin(?), hence the limit is 1. Now both limits in the proof I gave in my initial post can be determined without the use of any other properties other than what I gave above and the Pythagorean identity 1 - cos²(?) = sin²(?).

From my understanding of the Taylor series for sine, it's necessary to take the derivative of the sine function to arrive at that series.

December 23rd, 2013, 05:45 AM   #5
Math Team

Joined: Sep 2007

Posts: 2,409
Thanks: 6

Re: How to prove this question?

Quote:
Originally Posted by greg1313
Quote:
 Originally Posted by HallsofIvy But there are logical difficulties with defining "arc length" for a unit circle before you have defined sine and cosine . . .
Such as? If we let one radian be the angle subtending the arc length equivalent to the radius of the unit circle, then arc length = r? with r being the radius of the circle and ? being the angle subtending the arc.
How do you prove that? If you try to prove that, for a circle, arclength= r?, you wind up having to do an arclength integral which, eventually, requires using sine and cosine, either directly or indirectly.

Quote:
 A geometric interpretation of the limit sin(?)/? can now be given: as ? approaches 0, sin(?) approaches the arc length of ? radians. This can be visualized as the arc length getting closer and closer to a straight line, that is, closer and closer to sin(?), hence the limit is 1. Now both limits in the proof I gave in my initial post can be determined without the use of any other properties other than what I gave above and the Pythagorean identity 1 - cos²(?) = sin²(?). From my understanding of the Taylor series for sine, it's necessary to take the derivative of the sine function to arrive at that series.
I did NOT say "use the Taylor series", I said define sine and cosine to be those given series. You do not have to "arrive" at them.

December 23rd, 2013, 05:58 AM   #6
Global Moderator

Joined: Oct 2008
From: London, Ontario, Canada - The Forest City

Posts: 7,898
Thanks: 1093

Math Focus: Elementary mathematics and beyond
Re: How to prove this question?

Quote:
 Originally Posted by HallsofIvy How do you prove that?
As you say, it's a definition. And the Taylor series is defined in terms of the derivative of the function for which the series is given.

 Tags prove, question

 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post 450081592 Applied Math 1 September 15th, 2010 04:35 AM 450081592 Calculus 0 September 14th, 2010 06:26 PM 450081592 Linear Algebra 1 February 25th, 2010 03:16 AM igalep132 Calculus 8 February 7th, 2008 05:36 AM jiasyuen Algebra 0 December 31st, 1969 04:00 PM

 Contact - Home - Forums - Cryptocurrency Forum - Top