All Questions
41
questions
7
votes
3
answers
157
views
History of the general formula for linearising $\cos^n(x)$
I was wondering where the formula:
$$\cos^n(x)=\sum_{k=0}^n\frac{n!}{k!(n-k)!}\cos(x(2k-n))$$
Was first originally published. I accidentally derived it a few days ago and was wondering where it was ...
- 584
18
votes
2
answers
497
views
Prove $\sum^{n+1}_{j=1}\left|\cos\left(j\cdot x\right)\right|\geqslant \frac n4$
How do you prove that $\displaystyle\sum^{n+1}_{j=1}\left|\cos\left(j\cdot x\right)\right|\geqslant \dfrac{n}{4}$, where $x\in\mathbb{R}$?
I tried mathematical induction, but it doesn't work. I also ...
- 405
0
votes
5
answers
202
views
Evaluating $\sum_{k=1}^{10}\left(\sin\frac{2k\pi}{11}+i\cos\frac{2k\pi}{11}\right)$
Find the value of $$\sum_{k=1}^{10}\left(\sin\frac{2k\pi}{11}+i\cos\frac{2k\pi}{11}\right)$$
I have heard somewhere that this question can be done by $nth$ roots of unity or by vector algebra. But I'...
- 1,253
1
vote
1
answer
117
views
If $\beta=e^{i\frac{2\pi}7}$, and $\left|\sum_{r=0}^{3n-1}\beta^{2^r}\right|=4\sqrt2$, find $n$ [duplicate]
If $\beta=e^{i\frac{2\pi}7}$, and $\left|\sum_{r=0}^{3n-1}\beta^{2^r}\right|=4\sqrt2$, find $n$.
My Attempt:
$\beta+\beta^2+\beta^4+\beta^8+\beta^{16}+...=4\sqrt2$
This is no GP. Can we do something ...
- 8,338
-1
votes
2
answers
99
views
What trigonometric identities was used to get this? [closed]
I have this solution for a math question, but I didn't get how did we get from the line $3$ to $4$ and also from $4$ to $5$. I can understand that this is done through trigonometric identities. but ...
- 1
3
votes
1
answer
359
views
Finding the sum of $\cos{x}+2\cos{2x}+...+n\cos{nx}$ [duplicate]
I'm struggling to find the sum of $S_n=\cos{x}+2\cos{2x}+3\cos{3x}...+n\cos{nx}$
I know that for $z=e^{ix}$, $2\cos{nx}=z^n+\frac{1}{z^n}$.
So I've tried $2S_n=(z+2z^2+3z^3+...+nz^n)+(\frac{1}{z}+\...
- 1,381
4
votes
0
answers
128
views
How to solve $\sum_{n=-\infty}^\infty\frac{y^2}{[(x-n\pi)^2+y^2]^{3/2}}$?
I need to solve this sum:
$$\sum_{n=-\infty}^\infty\frac{y^2}{[(x-n\pi)^2+y^2]^{3/2}}.$$
Do you have any ideas for how I could do this?
I know that this sum:
$$\sum_{n=-\infty}^\infty\frac{y}{(x-n\pi)^...
- 1,027
-1
votes
2
answers
106
views
Summation from 0 to n of cos((k/n)2 pi) [duplicate]
Recently I came across this summation:
$$\sum_{k=0}^n\cos\Bigl(k\frac{2\pi}{n}\Bigr)$$ when I was trying to evaluate the following summation $$\sum_{k=0}^{n} z^k+z^{-k}$$ where $z \in \mathbb{C}, z=\...
- 36
1
vote
1
answer
194
views
Trigonometric Identities Using De Moivre's Theorem
I am familiar with solving trigonometric identities using De Moivre's Theorem, where only $\sin(x)$ and $\cos(x)$ terms are involved. But could not use it to solve identities involving other ratios. ...
2
votes
1
answer
86
views
Find the sum of infinite series $\cos{\frac{\pi}{3}}+\frac{\cos{\frac{2\pi}{3}}}{2}+..$
Find the sum of infinite series $$\cos{\dfrac{\pi}{3}}+\dfrac{\cos{\dfrac{2\pi}{3}}}{2}+\dfrac{\cos{\dfrac{3\pi}{3}}}{3}+\dfrac{\cos{\dfrac{4\pi}{3}}}{4}+\dots$$
I tried to convert it to $\mathrm{cis}$...
- 1,157
3
votes
4
answers
174
views
Evaluate $\frac{q}{1+q^2}+\frac{q^2}{1+q^4}+\frac{q^3}{1+q^6}$, where $q^7=1$ and $q\neq 1$.
Let $q$ be a complex number such that $q^7=1$ and $q\neq 1$. Evaluate $$\frac{q}{1+q^2}+\frac{q^2}{1+q^4}+\frac{q^3}{1+q^6}.$$
The given answer is $\frac{3}{2}$ or $-2$. But my answer is $\pm 2$.
...
- 373
2
votes
1
answer
95
views
Proving $\sum_{r=0}^{n-1} (-1)^r \cos^n{\left(\frac{r\pi}{n}\right)}=\frac{n}{2^{n-1}}$
Prove that $$\sum_{r=0}^{n-1} (-1)^r \cos^n{\left(\dfrac{r\pi}{n}\right)}=\dfrac{n}{2^{n-1}}$$
I proved the result using induction, however am more interested in finding the sum using complex numbers....
- 1,157
0
votes
1
answer
78
views
Summation of an Infinite Series involving Trigonometry
I came across this summation problem the other day and I am not quite sure how to approach it
$$S=\sum_{n=0}^{n=\infty}\frac{2^{n-1}}{3^{2n-2}}\sin\left(\frac{\pi}{3.2^{n-1}}\right)$$
My approach ...
- 1,292
1
vote
1
answer
307
views
How to solve sum of cos(kx) for the case cos(x)=1
I have the solution for $\sum_{k=1}^n \cos(kx)$:
\begin{align}
\sum_{k=1}^n \cos(kx) & = \Re\left(\sum_{k=1}^n e^{ikx}\right)\\
& = \Re\left(e^{ix} {e^{inx}-1 \over e^{ix}-1}\right) \\
& ...
- 9
0
votes
2
answers
85
views
Exponential double angle formula
My question is whether someone could provide a proof for the following identity:
$$ \frac{1 - e^{int}}{1 - e^{it}} = e^{i(n-1)t/2} \frac{\sin(nt/2)}{\sin(t/2)} $$
Motivation:
The left hand side is ...
- 3
2
votes
2
answers
97
views
How do I prove that $\sum_{i=1}^m \cos^2\left(\frac{2\pi i}{m}\right) = \sum_{i=1}^m \sin^2\left(\frac{2\pi i}{m}\right) = \frac{m}{2}$? [duplicate]
I haven't had any good ideas nor found any helpful identities so far, so I'd appreciate some help.
Also, here $m > 2$.
Update: Thanks to the hints and to this previous post I managed to get to ...
- 6,833
0
votes
1
answer
74
views
Trigonometry and Complex Numbers with Series [duplicate]
The number
$$\text{cis}75^\circ + \text{cis}83^\circ + \text{cis}91^\circ + \dots + \text{cis}147^\circ$$ is expressed in the form $r \, \text{cis } \theta,$ where $0 \le \theta < 360^\circ$. ...
- 61
6
votes
2
answers
2k
views
Prove that $\cos \frac{2\pi}{2n+1}+\cos \frac{4\pi}{2n+1}+\cos \frac{6\pi}{2n+1}+...+\cos \frac{2n\pi}{2n+1}=\frac{-1}{2}$
Prove that $$\cos \frac{2\pi}{2n+1}+\cos \frac{4\pi}{2n+1}+\cos \frac{6\pi}{2n+1}+...+\cos \frac{2n\pi}{2n+1}=\frac{-1}{2}$$
My attempt,
Let an equation $x^{2n+1}-1=0$, which has roots $$\cos \frac{...
- 7,798
1
vote
1
answer
2k
views
Find the sum of a geometric progression involving cos using complex numbers and proof trigonometric formula
I got the following problem:
I need to prove - using complex numbers:
$\sum_{t=0}^n \cos(tb) = \frac{\cos\frac{nb}{2}\sin\frac{nb+b}{2}}{\sin\frac{b}{2}}$
Ok so what I came up with so far:
we know ...
- 537
1
vote
2
answers
46
views
Using Euler and polynomials
I want to show that $\sum_{k=-N}^{N}e^{ikx}=\frac{\sin((N+\frac{1}{2})x)}{\sin(\frac{x}{2})}$ for $N\in \mathbb{N}$
Any tips on how to proceed?
I tried doing it in two ways:
First using the sum of ...
- 1,394
1
vote
3
answers
106
views
If $n>3$ prove that $\sum_{k=0}^{n-1} (k-n)\cos\frac{2k\pi}{n}=\frac{n}{2}$.
Do you have any ideas on this IIT exercise?
If $n>3$ is an integer, prove that
$$\sum_{k=0}^{n-1} (k-n)\cos(2kπ /n) = n/2$$
In my attempt, I have considered
$$z=cis(2kπ/n), k=[1, 2,..., n-1]...
- 514
1
vote
0
answers
487
views
A generalised formula for $\cos {n\theta}$ in terms of powers of $\cosθ$ using De Moivre's Theroem
I am trying to generalise a formula for $\cos{nθ}$ in terms of powers of $\cosθ$ using De Moivre's Theorem for a high school assignment.
The equations are here:
I was wondering if letting $m= ⌊n/2⌋-j+...
- 101
4
votes
1
answer
263
views
How to calculate $\sum_{k=0}^n a^k\sin(kx)$?
I tried to evaluate
$$
\sum_{k=0}^n a^k\sin(kx)
$$
using complex numbers but it didn't work... Any hint?
$a$ and $x$ are real numbers.
- 145
0
votes
1
answer
104
views
How to prove this quasi-geometric trigonometric series identity without induction
$$\frac{2}{\sin{x}}\sum_{r=1}^{n-1} \sin{rx}\cos{[(n-r)y]} \equiv \frac{\cos{(nx)}-\cos{(ny)}}{\cos{x}-\cos{y}} - \frac{\sin{(nx)}}{\sin{x}}$$
The identity can be tediously proven using the Axiom of ...
- 4,394
7
votes
2
answers
305
views
Bounding a sum involving a $\Re((z\zeta)^N)$ term
This is a follow up to this question. Any help would be very much appreciated.
Let $k\in\mathbb{N}$ be odd and $N\in\mathbb{N}$. You may assume that $N>k^2/4$ or some other $N>ak^2$.
Let $\...
- 7,789
2
votes
2
answers
185
views
Simplifying this (perhaps) real expression containing roots of unity
Let $k\in\mathbb{N}$ be odd and $N\in\mathbb{N}$. You may assume that $N>k^2/4$ although I don't think that is relevant.
Let $\zeta:=\exp(2\pi i/k)$ and $\alpha_v:=\zeta^v+\zeta^{-v}+\zeta^{-1}$.
...
- 7,789
7
votes
3
answers
3k
views
Cotangent summation (proof)
How to sum up this thing, i tried it with complex number getting nowhere so please help me with this,$$\sum_{k=0}^{n-1}\cot\left(x+\frac{k\pi}{n}\right)=n\cot(nx)$$
- 164
1
vote
2
answers
217
views
Summing up trigonometric series [duplicate]
By considering $\sum_{r=1}^n z^{2r-1}$ where z= $\cos\theta + i\sin\theta$, show that if $\sin\theta$ $\neq$ 0, $$\sum_{r=1}^n \sin(2r-1)\theta=\frac{\sin^2n\theta}{\sin\theta}$$
I couldn't solve ...
- 346
1
vote
2
answers
1k
views
About a binomial expansion of complex numbers
Prove that $$1+{n \choose 1}\cos x + {n \choose 2}\cos 2x+... \cos nx=(2 \cos\frac{x}{2})^n(\cos\frac{nx}{2})$$ given that
$$(1+\cos x+i\sin x)^n=(2\cos\frac{x}{2})^n(\cos\frac{nx}{2}+i\sin\frac{nx}{2}...
- 37
0
votes
2
answers
78
views
find the coefficient
If $n$ is an odd natural number, and $\sin(n\theta) = \Sigma_{r=0}^{n} b_r \sin^r\theta$, then find $b_r$ in terms of $n$.
I have tried this using trigonometric expansion but unable to find solution ...
- 97