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Find: $$\lim_{n\to \infty}\left(\sin\frac{\pi}{2n}.\sin\frac{2\pi}{2n}.\sin\frac{3\pi}{2n}.....\sin\frac{(n-1)\pi}{2n}\right)^{1/n}$$

I was wondering,is there a formula for multiplication of sines,when angles are in arithmetic progression(likewise there is formula for summation of sines).I also tried this question by taking log of both sides and then applying L Hospital rule.But it is becoming messy.Can someone please assist me in solving this question?

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    $\begingroup$ Please avoid only-Latex titles, they are discouraged. $\endgroup$ – Vincenzo Oliva Aug 9 '15 at 13:30
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    $\begingroup$ @VincenzoOliva can you edit the title to something better? $\endgroup$ – djechlin Aug 9 '15 at 21:59
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We have $$\lim_{n\to\infty}\ln \left(\sin\frac{\pi}{2n}.\sin\frac{2\pi}{2n}.\sin\frac{3\pi}{2n}.....\sin\frac{(n-1)\pi}{2n}\right)^{1/n} =\lim_{n\to\infty}\sum_{k=1}^{n-1} \frac{1}{n}\ln \sin\left(\frac{k}{n}\cdot \frac{\pi}{2} \right) =\hspace{1cm}\int_{0}^{1} \ln \sin\frac{\pi t}{2} dt $$ Hence $$\lim_{n\to\infty} \left(\sin\frac{\pi}{2n}.\sin\frac{2\pi}{2n}.\sin\frac{3\pi}{2n}.....\sin\frac{(n-1)\pi}{2n}\right)^{1/n} =e^{\int_0^1\ln \sin\frac{\pi t}{2} dt} $$ But it can be calculated that $$\int_0^1\ln \sin\frac{\pi t}{2} dt =-\ln2$$ therefore $$\lim_{n\to\infty} \left(\sin\frac{\pi}{2n}.\sin\frac{2\pi}{2n}.\sin\frac{3\pi}{2n}.....\sin\frac{(n-1)\pi}{2n}\right)^{1/n} =e^{-\ln 2 }=\frac{1}{2} $$

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    $\begingroup$ It seems you have forgotten the $\sin$ in your integral. $\endgroup$ – Olivier Oloa Aug 9 '15 at 8:45
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    $\begingroup$ Yes you're right. I have just edit and now should be ok. $\endgroup$ – MotylaNogaTomkaMazura Aug 9 '15 at 8:48
  • $\begingroup$ OK. You may observe that $$ \int_{0}^{1} \ln \sin\frac{\pi t}{2} dt=-\ln 2.$$ $\endgroup$ – Olivier Oloa Aug 9 '15 at 8:52
  • $\begingroup$ But, how did calculate -ln2 $\endgroup$ – exilednick Aug 20 '15 at 7:18
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First we have \begin{align} \lim_{n\to \infty}\frac1{n}\ln{\left(\sin\frac{\pi}{2n}.\sin\frac{2\pi}{2n}\cdots\sin\frac{(n-1)\pi}{2n}\right)} &=\lim_{n\to \infty}\frac1{n}\sum\limits_{k=0}^{n-1}\ln{\left(\sin\frac{k\pi}{2n}\right)} \\ &=\frac{2}{\pi}\int_0^{\pi/2}\ln{\sin{x}}\:dx \\ &=\frac{2}{\pi}\int_0^{\pi/2}\ln{\cos{x}}\:dx\tag{$y=\pi/2-x$} \end{align} Let $I=\int_0^{\pi/2}\ln{\sin{x}}\:dx$. Then \begin{align} 2I&=\int_0^{\pi/2}\ln{\sin{x}}\:dx+\int_0^{\pi/2}\ln{\cos{x}}\:dx \\ &=\int_0^{\pi/2}\ln{\sin{x}\cos{x}}\:dx \\ &=\int_0^{\pi/2}(\ln{\sin{2x}}-\ln{2})\:dx \\ &=\int_0^{\pi/2}\ln{\sin{2x}}\:dx-\frac{\pi\ln{2}}{2} \\ &=\frac1{2}\int_0^{\pi}\ln{\sin{x}}\:dx-\frac{\pi\ln{2}}{2} \\ &=\frac1{2}\left(\int_0^{\pi/2}\ln{\sin{x}}\:dx+\int_{\pi/2}^{\pi}\ln{\sin{x}}\:dx\right)-\frac{\pi\ln{2}}{2} \\ &=\frac1{2}\left(\int_0^{\pi/2}\ln{\sin{x}}\:dx+\int_{0}^{\pi/2}\ln{\cos{x}}\:dx\right)-\frac{\pi\ln{2}}{2} \\ \end{align} So $I=-\dfrac{\pi\ln{2}}{2}$. And

\begin{align} \lim_{n\to \infty}\left(\sin\frac{\pi}{2n}.\sin\frac{2\pi}{2n}\cdots\sin\frac{(n-1)\pi}{2n}\right)^{1/n}&=\lim_{n\to \infty}e^{\frac1{n}\sum\limits_{k=0}^{n-1}\ln{\left(\sin\frac{k\pi}{2n}\right)}} \\ &=e^{\frac{2}{\pi}I}=e^{-\ln{2}} \\ &=\frac1{2} \end{align}

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$\bf{My\; Solution::}$ USing $\bf{n^{th}}$ root of Unity, $\displaystyle$ So Let $x= (1)^{\frac{1}{n}}\Rightarrow x^n - 1=0$

So $$x^n-1 = (x-1)\cdot (x-\alpha)\cdot (x-\alpha^2)\cdot.........(x-\alpha^{n-1})\;,$$

Where $$\displaystyle \alpha^r = \cos\left(\frac{2r\pi}{n}\right)+i\sin \left(\frac{2r\pi}{n}\right)\;,$$ and $r=0,1,2,3,.....(n-1)$

So $$\displaystyle \frac{x^n-1}{x-1} = (x-\alpha)\cdot (x-\alpha^2)\cdot.........(x-\alpha^{n-1})$$

Now Using The formula

$$\bullet\; x^n-1 = (x-1)\cdot \left(x^{n-1}+x^{n-2}+x^{n-3}+........+x^2+x+1\right)$$

So $$\displaystyle \left(x^{n-1}+x^{n-2}+x^{n-3}+........+x^2+x+1\right) = (x-\alpha)\cdot (x-\alpha^2)\cdot.........(x-\alpha^{n-1})$$

Now Put $x=1$ in above equation and taking Modulus on both side, We get

$$\displaystyle 1 = \left|(1-\alpha)\cdot (1-\alpha^2)\cdot.........(1-\alpha^{n-1})\right| = \left|(1-\alpha)\right|\cdot \left|(1-\alpha^2)\right|\cdot .........\left|(1-\alpha^{n-1})\right|$$

So we get

$$\displaystyle n= \left|1-\cos\left(\frac{2\pi}{n}\right)-i\sin \left(\frac{2\pi}{n}\right)\right|\cdot \left|1-\cos\left(\frac{4\pi}{n}\right)-i\sin \left(\frac{4\pi}{n}\right)\right|.........\left|1-\cos\left(\frac{2(n-1)\pi}{n}\right)-i\sin \left(\frac{2(n-1)\pi}{n}\right)\right|$$

So $$\displaystyle n = 2\sin\left(\frac{\pi}{n}\right)\cdot 2\sin\left(\frac{2\pi}{n} \right)....2\sin\left(\frac{(n-1)\pi}{n}\right)$$

So $$\displaystyle \sin\left(\frac{\pi}{n}\right)\cdot \sin\left(\frac{2\pi}{n} \right)....\sin\left(\frac{(n-1)\pi}{n}\right) = \frac{n}{2^{n-1}}$$

Now Replace $n\rightarrow 2n\;,$ We get

$$\displaystyle \sin\left(\frac{\pi}{2n}\right)\cdot \sin\left(\frac{2\pi}{2n} \right)....\sin\left(\frac{(n-1)\pi}{2n}\right) = \frac{2n}{2^{2n-1}}$$

So $$\displaystyle \lim_{n\rightarrow \infty}\left[\sin\left(\frac{\pi}{2n}\right)\cdot \sin\left(\frac{2\pi}{2n} \right)....\sin\left(\frac{(n-1)\pi}{2n}\right)\right]^{\frac{1}{n}} = \lim_{n\rightarrow \infty}\left(\frac{2n}{2^{2n-1}}\right)^{\frac{1}{n}}$$

$$\displaystyle = \frac{\lim_{n\rightarrow \infty}(2n)^{\frac{1}{n}}}{\lim_{n\rightarrow \infty}(2)^{2-\frac{1}{n}}} = \frac{1}{2^2}$$

For Calculation of $$\lim_{n\rightarrow \infty}(2n)^{\frac{1}{n}}$$

Let $$\displaystyle L=\lim_{n\rightarrow \infty}(2n)^{\frac{1}{n}}\Rightarrow \ln(L)=\lim_{n\rightarrow \infty}\frac{\ln(2n)}{n}$$

Now Using $$\bf{L,Hopital\; Rule}\;,$$ we get $$\ln(L)=0\Rightarrow L=e^{0}=1$$

I did not Understand Where I have Done Wrong, Plz anyone explain me, Thanks

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  • $\begingroup$ On the line after "So we get..." you've the product equal to $n$, but it looks like a straight rewrite of the earlier line, where its $1$. $\endgroup$ – user24142 Aug 9 '15 at 22:28

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