Trigonometric sequence limit

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sontea

#1 h Jan 11, 2016, 7:44 PM • 3 Y

Y by mma.sais, Adventure10, Mango247

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

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#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 $\prod_{k=1}^n \sin k < \left(\frac{3}{5}\right)^{\left \lfloor \frac{n}{3}\right \rfloor}\to 0$ .

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#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

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#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.

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#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

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#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.

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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.

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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.

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#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.

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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.

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

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

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