Consider z^n=1 , each root is \xi_k = \cos\frac {2k\pi} {n} + \mathrm i\sin\frac {2k\pi} {n} = \mathrm e^ {\mathrm i\frac {2k\pi} {n}}, k=0,1,2,...,n-1.

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I am working out a Fourier Series problem and I saw that the suggested solution used $\sin (n\pi) = 0$ and $\cos (n\pi)= (-1)^n$ to simply the expressions while finding the Fourier Coefficients $a_0$,...

You can geometrically show the result. Its equal to π π. Think about a regular polygon inscribed in a circle. The perimeter of the polygon is n ∗ s n ∗ s, where n n is the number of sides the polygon has, and s s is the side length of the polygon. s s is related to r r, the circles radius, by s 2 = r ∗ sin(2π 2n) s 2 = r ∗ sin (2 π ...

证明： 有n个根 由 (1)和 (2)得： 令: =2n-1⋅k=1n-1 sin kπn⋅k=1n-1-i⋅k=1n-1eikπn —吴崇试. 数学物理方法.3版.北京：北京大学出版社

2021年2月7日 · 你认为这个结论正确的前提只有一个：当n趋近于无穷。 因此，当你发现nπ也趋近于无穷，也满足此前提，于是直接套用上述结论，得出sinnπ的极限不存在的结论。 你错了。 不仅结论错了，而且前提也错了。 事实上，该结论有两个前提。 另一个前提是：n以1，2，3，……趋近于无穷。 nπ显然不满足此条件，所以不能直接套用这个结论。 从n=1开始，逐个代入上式，构成数列。 然后，发现该数列为常数数列，常数为0。 所以，该数列的极限为0。 即上式极限 …

There are two cases: n is even, write it as 2k and then you have \cos (k\pi) and \sin (k\pi) which you already know. n is odd, write it as 2k+1 and then you have \cos (k\pi+\frac\pi2) and \sin …

本文介绍了几个定理，描述了三类常见三角函数的无理性. 定理1 [1]： 其中 a_1,a_2,\dots,a_n 都是整数. 若 n 是奇数，则 a_n=0 ； n 为偶数，则 a_n= (-1)^ {\frac {n} {2}} . 证明： 根据 恒等式 \cos n\varphi+\cos (n-2)\varphi=2\cos (n-1)\varphi\cos\varphi ，即得 递推关系式. \cos n\varphi=2\cos (n-1)\varphi\cos\varphi-\cos (n-2)\varphi . 对 n 作数学归纳即得（1）.

What is sin (pi*n) ?

2016年3月22日 · You can check that sin nπ sin n π is near 0 0 (again not equal to 0 0 due to floating point precision) for "small" values of n n. This fact continues to hold for larger n n since sin sin is periodic. That is, sin x = sin(x + 2π) sin x = sin (x + 2 π) for all x ∈R x ∈ R.