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I know the definition of sine and cosine, but how were these formulas originally invented? I mean, how did people derive the power series for sine and cosine for the first time?

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In summary, the conversation discusses the origins of the sine and cosine formulas and their definitions. It is mentioned that historically, the original definitions were in terms of right triangles and the derivatives of sine and cosine were derived from those definitions. The Taylor's series for sine and cosine are also mentioned as a way to define these trigonometric functions. The conversation also discusses alternative definitions for sine and cosine, such as defining them as functions satisfying certain conditions. The geometry of the unit-circle definitions is also mentioned as a visual interpretation of these functions. The conversation concludes with a discussion about approximations of sine and cosine before the discovery of Taylor series.

- #1

- 302

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I know the definition of sine and cosine, but how were these formulas originally invented? I mean, how did people derive the power series for sine and cosine for the first time?

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- #2

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You can also define sine and cosine by:

"y= sin(x) is the function satisfying y"= -y for all x, with y(0)= 0, y'(0)= 1 and y= cos(x) is the function satisfying y"= -y for all x, with y(0)= 1, y'(0)= 0" and, personally, I prefer that definition. From that the Taylor's series about x= 0 immediately follow.

- #3

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(cos(x+dx)-cos(x))/dx = (cos(x)cos(dx)-sin(x)sin(dx)-cos(x))/dx = sin(x)sin(dx)/dx --> sin(x)

? (I'm sorry this is so ugly, I hope you understand what I mean. Is TeX broken?)

The only problem is that I have to argue that sin(dx) goes to zero linearly, but that seems to make sense geometrically.

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Edit:

Yes, LaTeX is broken at the moment :(

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So we have (1/2)cos(t)sin(t)<= (1/2)t<= (1/2)sin(t)/cos(t). Multiplying through by 2/sin(x), we have cos(t)<= t/sin(t)<= 1/cos(t) which, inverting, gives 1/cos(t)<= sin(t)/sin(t)<= cos(t). Since cos(t) is continuous and cos(0)= 1, taking the limit as t goes to 0, lim sin(t)/t= 1.

We also will need sin(x+y)= sin(x)cos(y)+ cos(x)sin(y) and sin(x-y)= sin(x)cos(y)- cos(x)sin(y). Adding those sin(x+y)- sin(x- y)= 2cos(x)sin(y). In particular, if we take A= x+y and B= x-y, then x= (A+ B)/2 and y= (A-B)/2 so that sin(A)- sin(B)= 2 cos((A+B)/2)sin((A-B)/2).

Now we can write sin(x+h)- sin(x)= 2cos((2x+h)/2)sin(h/2)= 2 cos(x+ h/2)sin(h/2)

Then [sin(x+h)- sin(x)]/h= 2 cos(x+h/2)sin(h/2)/h= cos(x+h/2)sin(h/2)/(h/2).

lim(h->0) [sin(x+h)- sin(x)]/y= lim(h->0) cos(x+ h/2) lim(h->0) sin(h/2)/(h/2). cos(x+ h/2) goes to cos(x) and sin(h/2)/(h/2) goes to 1 so that limit is cos(x): the derivative of sin(x) is cos(x).

- #6

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sin(x)=x-(x^3)/3+...

does this not go to zero linearly ? I.e. lim(x->0) sin(x) = lim(x->0) x.

Anyway, it is a nice proof and a nice way to derive sine and cosine that you have told me. What about before people knew Taylor series? I just tried to do some geometric approximations (approximating an arc by a straight line), and I got

cos(x)=1-1/2*x^2

but I could not get much further because the equations got too complicated ...

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