Is this a variant of Euler's Formula? Understanding e^iy = cosy + isiny

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Euler's formula states that e^ix = cosx + isinx, and the variant e^ikx = coskx + isinkx is derived by substituting y = kx. This substitution maintains the fundamental relationship of Euler's formula while adjusting for the frequency factor k. The formula remains a valid application of Euler's principles, demonstrating how the exponential function can represent oscillatory behavior in a more generalized form. Understanding this transformation clarifies the connection between complex exponentials and trigonometric functions. The discussion emphasizes the consistency of Euler's formula across different contexts.
SpartanG345
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Eulers formula says that e^ix = cosx + isinx

but in my textbook there's another formula its e^ikx = coskx + isinkx
i still can't figure out how they got that. Is this still eulers formula? and how do u get it in that form
 
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e^iy = cosy + isiny

Let y = kx

e^ikx = coskx + isinkx
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Essentially I just have this problem that I'm stuck on, on a sheet about complex numbers: Show that, for ##|r|<1,## $$1+r\cos(x)+r^2\cos(2x)+r^3\cos(3x)...=\frac{1-r\cos(x)}{1-2r\cos(x)+r^2}$$ My first thought was to express it as a geometric series, where the real part of the sum of the series would be the series you see above: $$1+re^{ix}+r^2e^{2ix}+r^3e^{3ix}...$$ The sum of this series is just: $$\frac{(re^{ix})^n-1}{re^{ix} - 1}$$ I'm having some trouble trying to figure out what to...
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