Trigonometric sequence limit

Art of Problem Solving

AoPS Online

College Math Topics in undergraduate and graduate studies

Topics in undergraduate and graduate studies

3 M G

BBookmark VNew Topic kLocked

College Math Topics in undergraduate and graduate studies

Topics in undergraduate and graduate studies

3 M G

BBookmark VNew Topic kLocked

No tags match your search

M

Trigonometric sequence limit

G H J

Reveal topic tags

L

G H BBookmark kLocked kLocked NReply

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

35 posts

sontea

#1 h Jan 11, 2016, 7:44 PM • 3 Y

Y by mma.sais, Adventure10, Mango247

Evaluate $\lim_{n\rightarrow\infty}\sin1\sin2...\sin n.$

This post has been edited 1 time. Last edited by sontea, Jan 11, 2016, 7:46 PM

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

3469 posts

#2 h Jan 11, 2016, 8:11 PM • 3 Y

Y by Tashm, Adventure10, Mango247

Out of any three consecutive numbers, at least one must fall at or below $\frac{\pi}{2}-1$ or above $\frac{\pi}{2}+1$ modulo $\pi$ . This means $\min \lbrace |\sin x|, |\sin (x+1)|, |\sin (x+2)| \rbrace \leq \cos 1<\frac{3}{5}$ , and therefore $\prod_{k=1}^n \sin k < \left(\frac{3}{5}\right)^{\left \lfloor \frac{n}{3}\right \rfloor}\to 0$ .

This post has been edited 1 time. Last edited by JoeBlow, Jan 11, 2016, 8:20 PM

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

5 posts

#3 h Jan 11, 2016, 8:36 PM • 2 Y

Y by Adventure10, Alphaamss

if you put $S_ {n}=sin(1)...sin(n)$ then you have $|S_ {n+1}|= |sin(n+1)||S_{n}|<|S_{n}|$ (n+1 can't be of the form $k\pi$ ). Then $|S_{n}|$ is decreasing positive thus convergent to a limit $l$ . if $l!=0$ then $lim \frac{S_{n+1}}{S_{n}}=1=sin(n+1)$ which is not possible as $sin(n+1)$ have no limit

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

3469 posts

#4 h Jan 11, 2016, 8:55 PM • 1 Y

Y by Adventure10

fermate88 wrote:

...which is not possible as $sin(n+1)$ have no limit

Sure, but proving this assertion is precisely the non-trivial content of the problem.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

5 posts

#5 h Jan 12, 2016, 3:47 PM • 1 Y

Y by Adventure10

suppose the sequence $sin(n)$ converge to $\alpha$ , if you put $x_{n}=sin(n)$ and $y_{n}=sin(n)$ then $y_{n}$ converge as well to $\beta$ .
and we have $\alpha^2+\beta^2=1$ .
In the other side $x_{n+1}=x_{n}sin(1)+y_{n}cos(1)$ $y_{n+1}=y_{n}sin(1)+x_{n}cos(1)$ , if you take the limit, you end up with the system:
$\alpha=\alpha sin(1)+\beta cos(1)$ and $\beta=\beta sin(1)+\alpha cos(1)$ , which a unique solution $(\alpha,\beta)=(0,0)$ , which is a contradiction with the first equation

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

3469 posts

#6 h Jan 12, 2016, 5:29 PM • 1 Y

Y by Adventure10

fermate88 wrote:

...if you put $x_{n}=sin(n)$ and $y_{n}=sin(n)$ then $y_{n}$ converge as well to $\beta$ .

Assuming you meant $y_n=\cos n$ , the claim is false. The convergence of one implies only the convergence of the other in absolute value. The recurrences you wrote are also not quite right.

Still, the proof idea is a good one and will work after some minor fixes. But there's a more fundamental fact here that has nothing to do with manipulating trigonometric identities: if $f$ is a continuous periodic function with irrational period, then $\lbrace f(n):n\in \mathbb{N}\rbrace$ is dense in $f(\mathbb{R})$ , and therefore $\lim_{n\to \infty}f(n)$ exists if and only if $f$ is constant. Proving this should require only basic definitions and the pigeonhole principle.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

35 posts

sontea

#7 h Jan 13, 2016, 7:45 AM • 2 Y

Y by JoeBlow, Adventure10

JoeBlow wrote:

Out of any three consecutive numbers, at least one must fall at or below $\frac{\pi}{2}-1$ or above $\frac{\pi}{2}+1$ modulo $\pi$ . This means $\min \lbrace |\sin x|, |\sin (x+1)|, |\sin (x+2)| \rbrace \leq \cos 1<\frac{3}{5}$ , and therefore $\prod_{k=1}^n \sin k < \left(\frac{3}{5}\right)^{\left \lfloor \frac{n}{3}\right \rfloor}\to 0$ .

I think the inequality $|\prod_{k=1}^n \sin k |<\prod\min \lbrace |\sin k|, |\sin (k+1)|, |\sin (k+2)| \rbrace$ , which you use in your aproach, is not true.

This post has been edited 1 time. Last edited by sontea, Jan 13, 2016, 7:46 AM

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

35 posts

sontea

#8 h Jan 13, 2016, 8:15 AM • 2 Y

Y by Adventure10, Mango247

JoeBlow wrote:

fermate88 wrote:

...which is not possible as $sin(n+1)$ have no limit

Sure, but proving this assertion is precisely the non-trivial content of the problem.

If $\sin =n$ have limit, that limit is finite.Since $\sin (n+2)-\sin n=2\sin 1\cos (n+1)$ we obtain $\cos n\rightarrow 0$ .
$\sin 2n=2\sin n\cos n \Rightarrow \sin 2n \rightarrow 0 \Rightarrow \sin n \rightarrow 0$ .
But $\lim(\cos^2n+\sin^2n)=1=\lim \sin^2n+\lim\cos^2n=0+0$ , false.

This post has been edited 1 time. Last edited by sontea, Jan 13, 2016, 8:16 AM

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

3469 posts

#9 h Jan 13, 2016, 7:33 PM • 2 Y

Y by Adventure10, Mango247

sontea wrote:

$|\prod_{k=1}^n \sin k |<\prod\min \lbrace |\sin k|, |\sin (k+1)|, |\sin (k+2)| \rbrace$ , which you use in your aproach, is not true.

The inequality you wrote is very trivially false insofar as it's always true that $a\geq \min \lbrace a,b,c \rbrace$ , regardless of what $a,b,c$ are. Luckily, that is not an inequality I am using anywhere; please think about the post more carefully.

This post has been edited 1 time. Last edited by JoeBlow, Jan 13, 2016, 7:34 PM

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

35 posts

sontea

#10 h Jan 14, 2016, 4:32 AM • 2 Y

Y by Adventure10, Mango247

JoeBlow wrote:

sontea wrote:

$|\prod_{k=1}^n \sin k |<\prod\min \lbrace |\sin k|, |\sin (k+1)|, |\sin (k+2)| \rbrace$ , which you use in your aproach, is not true.

The inequality you wrote is very trivially false insofar as it's always true that $a\geq \min \lbrace a,b,c \rbrace$ , regardless of what $a,b,c$ are. Luckily, that is not an inequality I am using anywhere; please think about the post more carefully.

A, now I understand. But, if $x_n\leq y_n\rightarrow 0$ is not true than $x_n\rightarrow 0$ .
Example $x_n=-1+\frac{1}{n}, y_n=\frac{1}{n}$ .
Finally, I undersand your aprouch, tkx.

This post has been edited 3 times. Last edited by sontea, Jan 14, 2016, 11:20 AM

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

134 posts

#12 h Apr 30, 2025, 3:33 AM

Y by

$\begin{align*} &\left| \sin x \sin 2x \right| = \left| 2 \sin^2 x \cos x \right| = \left| 2( 1-\cos^2 x) \cos x \right| \\ &= \sqrt{2} \sqrt{( 1-\cos^2 x)^2 \, 2\cos^2 x} \\ &\le \sqrt{2} \sqrt{ \left( \frac{( 1-\cos^2 x) +( 1-\cos^2 x) + 2\cos^2 x}{3} \right)^3 }\\ &=\sqrt{ \frac{16}{27}} <1 \end{align*}$

Z K Y

N Quick Reply

G

H

=

© 2025 AoPS Incorporated

a