55
votes
A conjectural trigonometric identity
Let $T_n$ be the $n$-th Chebyshev polynomial, so that
$$T_n(x)-1=\prod_{j=1}^n(x-\cos 2\pi j/n).$$
Taking a logarithmic derivative we have
$$\sum_{j=0}^{n-1}\frac{1}{x-\cos 2\pi j/n}=\frac{T_n'(x)}{...
- 28.2k
42
votes
Accepted
Boundedness of sum of sin(sin(n))
This is true. It doesn't have much to do with the details of $\sin(\sin(\ ))$. Rather:
Theorem Let $f: \mathbb{R} \to \mathbb{R}$ be any smooth, $2 \pi$-periodic function with $\int_{z=0}^{2 \pi} f(z) ...
- 156k
35
votes
A trigonometric equation: how hard could it be?
I can't prove the identity, but I found that it implies a rather strange Dirichlet character sum identity, so I am recording the implication here in case anyone recognizes the latter.
I will ...
- 114k
31
votes
Accepted
A trigonometric equation: how hard could it be?
In the comments, GH from MO shows that it suffices to prove that
$$
\sum_{k=1}^{n-1} (\csc(2kx)+\tan(\pi/6+kx)-\tan(\pi/6+2kx))=0,\tag{$\ast$}
$$
and suggests replacing $\tan$ with $\cot$ using $\pi/6=...
- 1,538
30
votes
A conjectural trigonometric identity
Denote $\omega=e^{2\pi i/n}$, then $\Omega=\{\omega^k:k=0,\ldots,n-1\}$ is the set of roots of the polynomial $f(t):=t^n-1$, $-\Omega=\{-\omega^k:k=0,\ldots,n-1\}$ the set of roots of $g(t):=t^n+1$. ...
- 109k
17
votes
Accepted
Better trigonometrical inequalities for $\zeta(s)$?
Assuming the $b_i$ are all distinct (or at least non-zero for $i \neq 0$), this is not possible. (Otherwise there are trivial examples, e.g. $1 + 2 \cos(0 \theta)+ \cos(0 \theta) \geq 0$ or $1 + 4 \...
- 114k
16
votes
Closed formula for sine powers
You are looking at the sum of $m$-th powers of the imaginary parts of $n$-th roots of $-1.$ First, note that these can be expressed as polynomials in elementary symmetric functions (of the imaginary ...
- 96.4k
16
votes
Integer solutions of $2\cos\left(\frac{p\pi}n\right)+2\cos\left(\frac{q\pi}n\right)+4\cos\left(\frac{p\pi}n\right)\cos\left(\frac{q\pi}n\right)=1$
Yes, these are the only solutions. Let $\alpha = e^{p \pi i/n}$ and $\beta = e^{q \pi i/n}$. So the equation is
$$4 \left( \frac{\alpha+\alpha^{-1}}{2} \right) \left( \frac{\beta+\beta^{-1}}{2} \...
- 156k
16
votes
Trigonometric inequality
There is a short way to take out trigonometry and remain with the sum of $\pm 1$, similar (but not immediately equivalent, see later) to these appearing in the Christophe Leuridan and Alexander ...
- 109k
14
votes
Accepted
Trigonometric inequality
I will continue from the point where Christophe Leuridan's answer stopped. Denote our sum $S$, then
$$
S=2p\sum_{n=1}^q \frac{\sin(2\pi np/q)}{(1+\cos(2πnp/q))\sin(2\pi n/q)}.
$$
For $n=q$ this is ...
- 7,193
12
votes
Trigonometric inequality
I give a way that may work for odd numbers. This is too long for a comment.
First, the quantity
$$\cos\Big(\frac{2m\pi}{p}\Big)+\cos\Big(\frac{2n\pi}{q}\Big)
= 2\cos\Big(\pi\Big(\frac{m}{p}+\frac{n}{...
- 3,808
11
votes
A trigonometric equation: how hard could it be?
This is not an answer either. Instead I wish to address Nate River's question. In addition, it might help someone take advantage of this approach to solve the problem.
I run into the trigonometric ...
- 43.2k
10
votes
Boundedness of sum of sin(sin(n))
I think one can use similar tricks as in the accepted answer at Is the series $\sum_n|\sin n|^n/n$ convergent?.
To start with, I don't think one can hope for a constant bound $C$ as you hoped. Indeed,
...
- 10.8k
9
votes
Accepted
A conjectural trigonometric identity
The following argument is very short, but bit tricky, so I remain it along with the previous answer.
Actually this sum quickly factorizes: using $\cos (x+y)+\cos (x-y)=2\cos x\cos y$ and denoting $(j+...
- 109k
8
votes
Asymptotic behavior of a certain trigonometric partial sum
The desired inequality should be true iff
$$
c < c_0 := (r - \sqrt{r^2-1})^2
\quad\ \text{where} \quad\
r = \frac{|a|}{2b}
$$
(NB the hypotheses $b>0$ and $a < -2b$ imply $r>1$, so $0 < ...
- 79.9k
8
votes
A problem in additive combinatorics
I have nothing to add to the Fourier-type approach suggested in the question, but for those curious, thought it useful to outline the combinatorial solution to the problem that I know (I believe this ...
- 7,003
8
votes
Accepted
The complex trigonometric function degenerates to the positive integer
Following Johann Cigler's suggestion, set $q=e^{\frac{i\pi}{N}}$. We will need the two evaluations
$$\prod_{n=1}^{2N-1}(1-q^n)=\left.\frac{x^{2N}-1}{x-1}\right|_{x=1}=2N \tag{1}$$
$$\prod_{n=1}^{N-1}(...
- 85.6k
7
votes
coordinate free foundations of trigonometry
Trigonometric functions do not belong to geometry. Neither does the "measurement of angles" by real numbers.
They belong to analysis. This fact is discussed in detail in the book of Dieudonne, Linear ...
- 91.8k
7
votes
positive sum of sines
Let me begin by restating your conjecture. Consider the sum
$$S(x):=\sum_{j=1}^nb_j\sin jx,\quad b_n=1,\quad b_j\in\{0,1\}.$$
Then $S(x)\geq 0$ on $[0,\pi]$ if and only if $b_j=1$ for all odd $j$ and $...
- 91.8k
6
votes
The first zero-crossing of a combination of sines
That one cannot guarantee a zero on $(0,\pi/\delta]$ or on any
interval $(0,c]$ with $c$ independent on $n$ follows from the example
of B. Logan, (Theorem 5.5.1 of his thesis Properties of high-pass ...
- 91.8k
6
votes
Accepted
Inequality involving sine and cosine
I am afraid it is false. Take $F=0$, $A=C$, $E=2A$, $\mu=1/2$. Then we are given $\cos B+\cos D=2\cos A$ and should prove $\cos B/2+\cos D/2\geqslant 2\cos A/2$. But if $\cos B=x$, then $\cos B/2=\...
- 109k
5
votes
Better trigonometrical inequalities for $\zeta(s)$?
this is an answer to the question as originally posed, without the additional conditions on $a_0$ and $b_0$
for example,
$$6+\cos \theta+2 \cos 2 \theta+3 \cos 3 \theta\geq 0,$$
or more ...
- 188k
5
votes
Closed formula for sine powers
Letting $\xi=e^{\frac{2\pi i}{2n}}$, a primitive root of unity $z^{2n}=1$. Then, $\sin\left(\frac{\pi k}n\right)=\frac{\xi^{k}-\xi^{-k}}{2i}$ and hence
\begin{align}
\sum_{k=0}^{n-1}\sin^m\left(\frac{\...
- 43.2k
5
votes
Smallest regular $m$-gon covering a regular $n$-gon
Let $p_i = [x_i,y_i]$ be the vertices of the $n$-gon. To minimize the $m$-gon covering it, we minimize $r$ subject to the constraints
$$ \cos(\theta + 2 \pi j/m) (x_i - x) + \sin(\theta+2\pi j/m) (...
- 54.2k
4
votes
Distribution of distances of successive zeros of $f(x)={\rm cos}(x)+{\rm cos}(\alpha x)+{\rm cos}(\beta x)$
The peaks arise from envelopes that arise when the function value nearly repeats. The peaks are at the roots of the implicit equation $\pm {\rm cos}(x)\pm {\rm cos}(\alpha x)\pm {\rm cos}(\beta x)=0$ ...
4
votes
Accepted
Something interesting about the quintic $x^5 + x^4 - 4 x^3 - 3 x^2 + 3 x + 1=0$ and its cousins
I think the depressed quintic in the question is what Emma Lehmer called the reduced quintic in her paper, The quintic character of 2 and 3, Duke Math. J. Volume 18, Number 1 (1951), 11-18, MR0040338. ...
- 39.9k
4
votes
Identities for Chebyshev polynomials of the second kind
Mathematica evaluates
$$S_n=\sum_{k=0}^n\sum_{m=0}^{2k}(-2)^m\binom{2(n-k)}{n-k}\binom{2k}{k}\binom{2k}{m}[\cos{\phi}]^mU_m(\cos{\phi})=$$
$$=\frac{(-1)^n 4^n\sqrt\pi}{n!\,\Gamma(\tfrac{1}{2}-n)\sin\...
- 188k
4
votes
Show that $(\sum_{k=1}^{n}x_{k}\cos{k})^2+(\sum_{k=1}^{n}x_{k}\sin{k})^2\le (2+\frac{n}{4})\sum_{k=1}^{n}x^2_{k}$
It is a bit long for a comment.
Your question is about the matrix $A=(\cos((i-j)))_{i,j=1\ldots n}$, specifically, the maximum of the quadratic form $q(x)=(Ax,x)$ on the subset $M_+$ of the unit ...
- 16.9k
4
votes
Accepted
The cotangent sum $\sum_{k=0}^{n-1}(-1)^k\cot\Big(\frac{\pi}{4n}(2k+1)\Big)=n$
The expression under the limit sign in question is just a Riemann sum for
$$\int_0^{\pi/2} \frac12\,\Big(\frac1x-\cot x\Big)\,dx=\frac12\,\ln\frac\pi2,$$
which therefore is the value of the limit.
- 128k
4
votes
Sum of $\sin$ when angles shrink by $1/n$
For large $n$ you may approximate the sum by an integral, which gives
$$\sum_{k=1}^{n} k \sin (x/k)-nx\simeq \int_0^\infty\bigl(k\sin(x/k)-k\bigr)\,dk=-\tfrac{1}{4}\pi x^2.$$
The plot compares the ...
- 188k
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