Iterated trigonometric differentiation

[Orion1]

These are some equations that I recently developed and submitting for review.

Evaluations?, comments?

Iterated trigonometric differentiation:
$$\frac{d^n}{dx^n} \sin x = \sin \left(x + \frac{n \pi}{2} \right)$$

$$\frac{d^n}{dx^n} \cos x = \cos \left(x + \frac{n \pi}{2} \right)$$

Iterated trigonometric integration:
$$(I^n \sin)(x) = \sin \left( x - \frac{n \pi}{2} \right)$$
$$(I^n \cos)(x) = \cos \left( x - \frac{n \pi}{2} \right)$$
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[vkgoku2012]

How do you understand math like that what does that all mean.

 

[gnomedt]

Looks fine to me. Are you trying to find the half-derivative of trig functions?

 

[Orion1]

How do you understand math like that what does that all mean?

Iteration is a 'repeated' process used in higher order mathematics, for example:
$$\frac{d^2}{dx^2} x^3 = \frac{d}{dx} \left( \frac{d}{dx} x^3 \right) = \frac{d}{dx} \left( 3x^2 \right) = 6x$$

Therefore, the second iterated derivative of this function is:
$$\frac{d^2}{dx^2} x^3 = 6x$$

The derivative is repeatedly derived twice, or 'iterated' twice.

Are you trying to find the half-derivative of trig functions?

Negative, are you referring to 'fractional' calculus?
Is this a demonstration of a half-derivative trig function?
$$\frac{d^{\frac{1}{2}}}{dx^{\frac{1}{2}}} \sin x = \sin \left(x + \frac{\pi}{4} \right)$$
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