- #1
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How does...
[tex]e^{\frac{1}{2} i n x} = \sin{ \frac{1}{2} n x}[/tex]
...where n is any positive integer and x is any angle.
I know about de Moivre's Theorem, but that can't be deduced from there.
There is also brackets around it, with some sort of greek letter on the outside. Looks like a vertheta or something. I know this doesn't help much, but I just wanted you to know. The sign is not present after the equal sign.
It would make sense if the sign (greek letter) is used to get rid of imaginary numbers somehow. Also, x would have to be one radian measure so that cos would always be eliminated. Because cos(pi n 1/2) will always be zero.
Maybe it doesn't even use de Moivre's Theorem.
I'm clueless.
Maybe the greek letter represents 1/i and therefore gets rid of the i.
Honestly, it doesn't even say what x is, it is actually that circle with the line across so that should represent any angle.
Can someone help me out here?
[tex]e^{\frac{1}{2} i n x} = \sin{ \frac{1}{2} n x}[/tex]
...where n is any positive integer and x is any angle.
I know about de Moivre's Theorem, but that can't be deduced from there.
There is also brackets around it, with some sort of greek letter on the outside. Looks like a vertheta or something. I know this doesn't help much, but I just wanted you to know. The sign is not present after the equal sign.
It would make sense if the sign (greek letter) is used to get rid of imaginary numbers somehow. Also, x would have to be one radian measure so that cos would always be eliminated. Because cos(pi n 1/2) will always be zero.
Maybe it doesn't even use de Moivre's Theorem.
I'm clueless.
Maybe the greek letter represents 1/i and therefore gets rid of the i.
Honestly, it doesn't even say what x is, it is actually that circle with the line across so that should represent any angle.
Can someone help me out here?