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December 21st, 2013, 06:27 PM   #1
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How to prove this question?

Prove that d(sin(x))/dx=cos(x),
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December 21st, 2013, 07:05 PM   #2
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Re: How to prove this question?







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December 21st, 2013, 08:03 PM   #3
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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
and


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
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December 22nd, 2013, 09:49 AM   #4
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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 - cosē(?) = sinē(?).

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.
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December 23rd, 2013, 05:45 AM   #5
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Re: How to prove this question?

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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 - cosē(?) = sinē(?).

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.
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December 23rd, 2013, 05:58 AM   #6
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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.
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