- #1

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Find [itex]\displaystyle\int \cos x \ \mathrm{d}x[/itex] without using the fact that the derivative of sine is cosine.

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- Thread starter ahaanomegas
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- #1

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Find [itex]\displaystyle\int \cos x \ \mathrm{d}x[/itex] without using the fact that the derivative of sine is cosine.

- #2

cepheid

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Find [itex]\displaystyle\int \cos x \ \mathrm{d}x[/itex] without using the fact that the derivative of sine is cosine.

How about using the Taylor series expansion of the cosine function?[tex] \int \cos x\,dx = \int \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{(2n)!}\,dx[/tex][tex] = \sum_{n=0}^{\infty} \int \frac{(-1)^n x^{2n}}{(2n)!}\,dx [/tex][tex] = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n)!}\frac{1}{2n+1}x^{2n+1}[/tex]Now, (2n+1)*(2n)! = (2n+1)! so that this becomes[tex]\sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{(2n+1)!}[/tex]Now all you have to do is "notice" that this result is the Taylor series expansion of a sine function.

I just realized that you'd probably have to know the derivative of cosine in order to derive its Taylor series expansion in the first place, so I'm not sure whether you are satisfied with this answer.

- #3

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Find [itex]\displaystyle\int \cos x \ \mathrm{d}x[/itex] without using the fact that the derivative of sine is cosine.

I think this is not possible. Maybe your friend considers

[tex]\sin\theta=\frac{{e}^{i\theta}-{e}^{-i\theta}}{2i}[/tex]

[tex]\cos\theta=\frac{{e}^{i\theta}+{e}^{-i\theta}}{2}[/tex]

to be definitions of sine and cosine? But sine and cosine are defined to y/r and x/r respectively.

- #4

cepheid

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I think this is not possible. Maybe your friend considers

[tex]\sin\theta=\frac{{e}^{i\theta}-{e}^{-i\theta}}{2i}[/tex]

[tex]\cos\theta=\frac{{e}^{i\theta}+{e}^{-i\theta}}{2}[/tex]

to be definitions of sine and cosine?

Assuming we're allowed that definition of cosine, then that also works.

But sine and cosine are defined to y/r and x/r respectively.

I'm not sure I understand your statement here. Can you clarify?

- #5

HallsofIvy

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Non-sense- because you haven't said what x, y, and rI think this is not possible. Maybe your friend considers

[tex]\sin\theta=\frac{{e}^{i\theta}-{e}^{-i\theta}}{2i}[/tex]

[tex]\cos\theta=\frac{{e}^{i\theta}+{e}^{-i\theta}}{2}[/tex]

to be definitions of sine and cosine? But sine and cosine are defined to y/r and x/r respectively.

You could also define cos(x) as [itex]\sum_{n=0}^\infty [(-1)^n/(2n)!]x^{2n}[/itex]

and sine(x) as [itex]\sum_{n=0}^\infty [(-1)^n/(2n+1)!]x^{2n+1}[/itex]

Or you can define cos(x) as the function satifying y''= -y with initiial conditions y(0)= 1, y'(0)= 0 and define sin(x) as the function satisfying y''= y with initial conditions y(0)= 0, y'(0)= 1.

- #6

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I'm not sure I understand your statement here. Can you clarify?

You haven't said what x, y, and rare.

If you draw a right-angled triangle, then x is the adjacent side, y is the opposite side and r is the hypotenuse.

- #7

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Non-sense- because you haven't said what x, y, and rare. There are, in fact, many different ways todefinesine and cosine. The formulas given above are perfectly good as definitions.

You could also define cos(x) as [itex]\sum_{n=0}^\infty [(-1)^n/(2n)!]x^{2n}[/itex]

and sine(x) as [itex]\sum_{n=0}^\infty [(-1)^n/(2n+1)!]x^{2n+1}[/itex]

Yes but I am considering the geometric definition.

define sin(x) as the function satisfying y''= y with initial conditions y(0)= 0, y'(0)= 1.

Isn't the second derivative of sin(x) is -sin(x) ?

- #8

Mark44

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Yes. I'm pretty sure HallsOfIvy meant y'' + y = 0 as the differential equation.Isn't the second derivative of sin(x) is -sin(x) ?

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