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4 votes
1 answer
108 views

Clean proof for trigonometry identity? I know what the answer is, but I feel like there should be like a $1$-$2$ liner to compute this

Fix $j,k$ with $0 \leq j,k \leq N$. If $j+k$ is even, (i.e. if $j,k$ have same parity), then $$ \sum_{n=1}^{N-1} \cos\left(\frac{j\pi}Nn\right)\sin\left(\frac{k \pi}Nn\right) = 0 $$ and $$ \sum_{n=0}^...
sweetpotato's user avatar
3 votes
1 answer
123 views

How should I prove that: $\sum_{i=1} ^{n}(\sin(\frac{i\pi}{n}))^2=\frac{n}{2}$

$$\sum_{i=1} ^{n}\Big(\sin\big(\frac{i\pi}{n}\big)\Big)^2=\frac{n}{2}$$ An interesting conclusion and checked for validity...holds for $n\geq 2$, but yet do not know how to prove it. Are there any ...
Amber Xue's user avatar
  • 345
3 votes
1 answer
383 views

How prove that $\frac{1}{\sin^2\frac{\pi}{2n}}+\frac{1}{\sin^2\frac{2\pi}{2n}}+\cdots+\frac{1}{\sin^2\frac{(n-1)\pi}{2n}} =\frac{2}{3}(n-1)(n+1)$ [duplicate]

How prove that sum $$\frac{1}{\sin^2\frac{\pi}{2n}}+\frac{1}{\sin^2\frac{2\pi}{2n}}+\cdots+\frac{1}{\sin^2\frac{(n-1)\pi}{2n}} =\frac{2}{3}(n-1)(n+1)$$
piteer's user avatar
  • 6,320
2 votes
1 answer
189 views

Evaluating $\sum_{n=1}^\infty \frac{\sin(nx)}{n}$ without integrating $\sum_{n=1}^\infty e^{nx}$

I am looking for alternative solutions for finding this sum $$\sum_{n=1}^\infty \frac{\sin(nx)}{n} $$ My solution proceeds by integrating $$\sum_{n=1}^\infty e^{nx}=\frac{e^{ix}}{1-e^{ix}}$$ With ...
Damn it My Foot's user avatar
2 votes
3 answers
256 views

Evaluate $\sum\limits_{r=1}^\infty(-1)^{r+1}\frac{\cos(2r-1)x}{2r-1}$

I would like to know how to evaluate $$\sum\limits_{r=1}^\infty(-1)^{r+1}\frac{\cos(2r-1)x}{2r-1}$$ There are a couple of issues I have with this. Firstly, depending on the value of $x$, it seems, at ...
A-Level Student's user avatar
2 votes
1 answer
224 views

Do there exist $a_k$ and $b_k$ so the equation $\sum\limits_{k=1}^{n} (a_k \sin(kx) + b_k \cos(kx)) = 0$ has no roots?

Do there exist real numbers $a_1, a_2, ..., a_n$ and $b_1, b_2, ..., b_n$ such that the equation $$\sum\limits_{k=1}^{n} (a_k \sin(kx) + b_k \cos(kx)) = 0$$ has no solutions?
PandaMan's user avatar
  • 3,269
2 votes
2 answers
185 views

Simplifying this (perhaps) real expression containing roots of unity

Let $k\in\mathbb{N}$ be odd and $N\in\mathbb{N}$. You may assume that $N>k^2/4$ although I don't think that is relevant. Let $\zeta:=\exp(2\pi i/k)$ and $\alpha_v:=\zeta^v+\zeta^{-v}+\zeta^{-1}$. ...
JP McCarthy's user avatar
  • 7,789
2 votes
1 answer
399 views

A finite sum of trigonometric functions

By taking real and imaginary parts in a suitable exponential equation, prove that $$\begin{align*} \frac1n\sum_{j=0}^{n-1}\cos\left(\frac{2\pi jk}{n}\right)&=\begin{cases} 1&\text{if } k \...
CatSensei's user avatar
2 votes
1 answer
276 views

$\displaystyle\sum_{k=0}^n \frac{\cos(k x)}{\cos^kx} = ?$

Prove that (not use induction) $\displaystyle\sum_{k=0}^n \frac{\cos(k x)}{\cos^kx} = \frac{1+(-1)^n}{2\cos^nx} + \dfrac{2\sin\big(\lfloor\frac{n+1}{2}\rfloor x\big) \cos\big(\lfloor\frac{n+2}{2}\...
hxthanh's user avatar
  • 1,520
1 vote
1 answer
183 views

Telescoping series: $\sum\limits_{n=1}^{∞}[\tan^{-1}(2n+1)-\tan^{-1}(2n-1)]$

In this question that was asked today the OP wrote that $$\begin{align}\sum\limits_{n=1}^{∞}[\arctan(2n+1)-\arctan(2n-1)]&=\arctan\infty-\arctan 1\\ &=\frac{\pi}{2}-\frac{\pi}{4}\\&=\frac{\...
A-Level Student's user avatar
1 vote
4 answers
396 views

Proving a function is continuous and periodic

Suppose we are given a function $$g\left ( x \right )= \sum_{n=1}^{\infty}\frac{\sin \left ( nx \right )}{10^{n}\sin \left ( x \right )},x\neq k\pi , k\in\mathbb{Z}$$ and $$g\left ( k\pi \right )=\...
Shemafied's user avatar
  • 485
1 vote
2 answers
18k views

Induction proof of the identity $\cos x+\cos(2x)+\cdots+\cos (nx) = \frac{\sin(\frac{nx}{2})\cos\frac{(n+1)x}{2}}{\sin(\frac{x}{2})}$ [duplicate]

Prove that:$$\cos x+\cos(2x)+\cdots+\cos (nx)=\frac{\sin(\frac{nx}{2})\cos\frac{(n+1)x}{2}}{\sin(\frac{x}{2})}.\ (1)$$ My attempt:$$\sin\left(\frac{x}{2}\right)\sum_{k=1}^{n}\cos{(kx)}$$$$=\sum_{k=1}^...
user180321's user avatar
1 vote
3 answers
233 views

Calculate $\sum_{n=1}^{\infty} \arctan\bigl(\frac{2\sqrt2}{n^2+1}\bigr) $

$$ \lim_{n \to\infty} \sum_{k=1}^{n} \arctan\frac{2\sqrt2}{k^2+1}= \lim_{n \to\infty} \sum_{k=1}^{n} \arctan\frac{(\sqrt{k^2+2}+\sqrt2)-\sqrt{k^2+2}-\sqrt2)}{(\sqrt{k^2+2}+\sqrt2)(\sqrt{k^2+2}-\sqrt2)+...
SADBOYS's user avatar
  • 1,219
1 vote
1 answer
120 views

Find the value of sum $\forall\:\:\alpha,\beta\in\mathbb{R}$

Evaluate the sum $\forall\:\:\alpha,\beta\in\mathbb{R}$ $$S=\sum_{n=1}^{\infty}\frac{\alpha^{n+1}-1}{n(n+1)}\sin\left(\frac{n\pi}{\beta}\right)$$ I rewrote this as $$S=\sum_{n=1}^{\infty}\left(\left(\...
MathStackexchangeIsMarvellous's user avatar
1 vote
1 answer
5k views

Summation of $\cos (2n-1) \theta$

By considering $\sum\limits_{n=1}^N z^{2n-1}$, where $z=e^{i\theta},$ show that $$ \sum\limits_{n=1}^N \cos{(2n-1)} \theta = \frac{\sin(2N\theta)}{2\sin\theta}, $$ where $\sin\theta\neq0$ I ...
M.S.E's user avatar
  • 1,927

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