Question 994566
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Can't.  It is only true at odd multiples of pi/2.  Without the lead coefficient of 2 in the numerator, then it works. Was that a typo?


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it

*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \