Forschungswebsite von Vyacheslav Gorchilin
Die Formeln. Mathematik. Trigonometrie
Nützliche Identitäten
$\sum\limits_{k=1}^n \sin(kx)$
$\sum \limits_{k=1}^{n}\cos(kx)$
$\sum \limits_{{k=1}}^{{n}} \sin {\big [}(2k-1)x{\big ]}$
$\sum \limits_{{k=1}}^{{n}}\cos {\big [}(2k-1)x{\big ]}$
$\sum \limits_{{k=1}}^{n}\cos \left({\frac{2\pi k}{2n+1}}\right)$
$\prod \limits_{k=1}^{n-1}\sin \left({\frac{k\pi }{n}}\right)$
$\prod \limits_{k=1}^{n}\sin \left({\frac{k\pi }{2(n+1)}}\right)$
$\prod \limits_{k=1}^{n}\sin \left({\frac{k\pi}{2n+1}}\right)$
$\prod \limits_{k=1}^{2n-1}\cos \left({\frac{k\pi}{n}}\right)$
$\prod \limits_{{k=0}}^{n}\cos \left(2^{k}x\right)$
$\prod \limits_{{k=0}}^{n}\cos \left({\frac{x}{2^{k}}}\right)$
$\prod \limits_{{k=1}}^{n}\cos \left({\frac{x}{2^{k}}}\right)$
$\prod \limits_{{k=0}}^{{\infty }}\cos \left({\frac{x}{2^{k}}}\right)$
$\prod \limits_{{k=1}}^{{\infty }}\cos \left({\frac{x}{2^{k}}}\right)$
$2^{{n-1}}\prod \limits_{{k=0}}^{{n-1}}\sin \left(x+{\frac{\pi k}{n}}\right)$
$(-1)^{n}\prod \limits_{{k=0}}^{{2n}}\mathtt{tg} \left(x+{\frac{\pi k}{2n+1}}\right)$
$\sum\limits_{k=1}^n \sin(kx) = \sin \left({\frac {n+1}{2}}x\right){\frac {\sin \left({\frac {nx}{2}}\right)}{\sin \left({\frac {x}{2}}\right)}}$
$\sum \limits_{k=1}^{n} \cos(kx) = \cos \left({\frac {n+1}{2}}x\right){\frac {\sin \left({\frac {nx}{2}}\right)}{\sin \left({\frac {x}{2}}\right)}}$
$\sum \limits_{{k=1}}^{{n}} \sin {\big [}(2k-1)x{\big ]} = {\frac{\sin ^{2}nx}{\sin x}}$
$\sum \limits_{{k=1}}^{{n}}\cos {\big [}(2k-1)x{\big ]} = {\frac{\sin 2nx}{2\sin x}}$
$\sum \limits_{{k=1}}^{n}\cos \left({\frac{2\pi k}{2n+1}}\right) = -{\frac 12}$
$\prod \limits_{k=1}^{n-1}\sin \left({\frac{k\pi}{n}}\right) = {\frac{n}{2^{n-1}}}$
$\prod \limits_{k=1}^{n}\sin \left({\frac{k\pi}{2(n+1)}}\right) = {\frac{\sqrt {n+1}}{2^{n}}}$
$\prod \limits_{k=1}^{n}\sin \left({\frac{k\pi}{2n+1}}\right) = {\frac{\sqrt {2n+1}}{2^{n}}}$
$\prod \limits_{k=1}^{2n-1}\cos \left({\frac{k\pi}{n}}\right) = {\frac {(-1)^{n}-1}{2^{2n-1}}}$
$\prod \limits_{{k=0}}^{n}\cos \left(2^{k}x\right) = {\frac{\sin \left(2^{{n+1}}x\right)}{2^{{n+1}}\sin x}}$
$\prod \limits_{{k=0}}^{n}\cos \left({\frac{x}{2^{k}}}\right) = {\frac{\sin 2x}{2^{{n+1}}\sin \left({\frac{x}{2^{n}}}\right)}}$
$\prod \limits_{{k=1}}^{n}\cos \left({\frac{x}{2^{k}}}\right) = {\frac{\sin x}{2^{{n}}\sin \left({\frac{x}{2^{n}}}\right)}}$
$\prod \limits_{{k=0}}^{{\infty }}\cos \left({\frac{x}{2^{k}}}\right) = {\frac{\sin 2x}{2x}}$
$\prod \limits_{{k=1}}^{{\infty }}\cos \left({\frac{x}{2^{k}}}\right) = {\frac{\sin x}{x}}$
$2^{{n-1}}\prod \limits_{{k=0}}^{{n-1}}\sin \left(x+{\frac{pi k}{n}}\right) = \sin nx$
$(-1)^{n}\prod \limits_{{k=0}}^{{2n}}\mathtt{tg} \left(x+{\frac{\pi k}{2n+1}}\right) = \mathtt{tg} {\big [}(2n+1)x{\big ]}$
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