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If if an exam I had to calculate the minimal polynomials of, say, $\sin(\frac{2 \pi}{5})$ or $\cos(\frac{2 \pi}{19})$, what would be the quickest way to do it? I can use the identity $e^{ \frac{2 i \pi}{n}} = \cos(\frac{2 \pi}{n}) + i\sin(\frac{ 2 \pi}{n})$ and raise to the power $n$, but this gets nasty in the $n = 19$ case... Thanks |
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$\cos(2\pi/19)$ has degree 9, and I doubt anyone would put it on an exam. $\sin(2\pi/5)$ is a bit more reasonable. Note that $\sin(4\pi/5)=2\sin(2\pi/5)\cos(2\pi/5)$, and also $\sin(4\pi/5)=\sin(\pi/5)$, and $\cos(2\pi/5)=1-2\sin^2(\pi/5)$, and with a few more identities like that you should be able to pull out a formula. Or use your idea in the form $2i\sin(2\pi/5)=z-z^{-1}$, where $z=e^{2\pi i/5}$, then, squaring, $-4\sin^2(2\pi/5)=z^2-2+z^{-2}$, $$-8i\sin^3(2\pi/5)=z^3-3z+3z^{-1}-z^{-3}=z^{-2}-3z+3z^{-1}-z^2$$ and a similar formula for the 4th power, and remember that $z^2+z+1+z^{-1}+z^{-2}=0$ |
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