(-1)^(1/x) = i*sin(pi/x) + cos(pi/x) where x>=4 is this equation new?

[Semjase]

I have no proof for it but just wondering if this equation is new?

 

 

(-1)^(1/x) = i*sin(pi/x) + cos(pi/x) where x>=4

 

I've tested this equation and Google calculator gives the correct answer

for every example.

Edited October 11, 2013 by Semjase

[John]

Well, if we consider Euler's formula and identity,

 

[math]e^{i\theta} = \cos \theta + i \sin \theta[/math]

 

and

 

[math]e^{i\pi} + 1= 0 \implies e^{i\pi} = -1[/math],

 

then your result follows pretty directly, since we have

 

[math]i \sin \frac{\pi}{x} + \cos \frac{\pi}{x} = e^{\frac{i\pi}{x}} = (e^{{i\pi}})^{\frac{1}{x}} = (-1)^{\frac{1}{x}}[/math].

Edited October 11, 2013 by John

[Semjase]

I'm just curious as to why that equation didn't work when plugged into

google calculator when x<4?

[John]

You'd probably have to ask Google about that.

 

As a particular example, Euler's identity itself is the case where x in your equation equals 1, and 1 < 4.

[Semjase]

You'd probably have to ask Google about that.

 

As a particular example, Euler's identity itself is the case where x in your equation equals 1, and 1 < 4.

 

Wolfram Alpha gave the correct result