Semjase Posted October 11, 2013 Share Posted October 11, 2013 (edited) 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 Link to comment Share on other sites More sharing options...
John Posted October 11, 2013 Share Posted October 11, 2013 (edited) 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 Link to comment Share on other sites More sharing options...
Semjase Posted October 11, 2013 Author Share Posted October 11, 2013 I'm just curious as to why that equation didn't work when plugged into google calculator when x<4? Link to comment Share on other sites More sharing options...
John Posted October 11, 2013 Share Posted October 11, 2013 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. Link to comment Share on other sites More sharing options...
Semjase Posted October 11, 2013 Author Share Posted October 11, 2013 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 Link to comment Share on other sites More sharing options...
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