Proving the product expansion of $sintheta$. Where did I go wrong?











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You can prove the expansion$$frac {sintheta}{theta}=left{1-left(frac {theta}{pi}right)^2right}left{1-left(frac {theta}{2pi}right)^2right}ldots$$By taking the expansion$$sin nphi=2^{n-1}sinphicosphileft(sin^2frac {pi}n-sin^2phiright)left(sin^2frac {2pi}{n}-sin^2phiright)ldots$$substituting $phi=theta/n$ and dividing by$$n=2^{n-1}sinfrac {pi}nsinfrac {2pi}nldotssinfrac {pi(n-1)}{n}$$





However, when I try, I always get zero as the answer. I started off with$$sintheta=2^{n-1}sinfrac {theta}ncosfrac {theta}nleft(sin^2frac {pi}n-sin^2frac {theta}nright)left(sin^2frac {2pi}n-sin^2frac {theta}nright)ldots$$And divided it to get$$frac {sintheta}n=frac {sinfrac {theta}ncosfrac {theta}nleft(sin^2frac {pi}n-sin^2frac {theta}nright)left(sin^2frac {2pi}n-sin^2frac {theta}nright)ldots}{sinfrac {theta}ncosfrac {theta}nldotssinfrac {pi(n-1)}n}$$However, when I multiply both sides by $n$ and take the limit as $n$ tends towards infinity, I get the expansion as$$begin{align*}sintheta & =thetaleft{frac 12-frac 12left(frac {theta}nright)^2right}left{frac 13-frac 13left(frac {theta}{2pi}right)^2right}ldots\\ & =thetaprodlimits_{k=1}^{infty}left{frac 1{k+1}-frac 1{k+1}left(frac {theta}{kpi}right)^2right}\ & =0end{align*}$$




Question: I'm trying to prove$$frac {sintheta}{theta}=prodlimits_{kgeq1}left{1-left(frac {theta}{kpi}right)^2right}$$So where did I go wrong?











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  • Do you want to prove the pointwise convergence for any $thetainmathbb{R}$ or $inmathbb{C}$? Or the uniform convergence over compact subsets of $mathbb{R}$ or $mathbb{C}$?
    – Jack D'Aurizio
    Aug 13 '17 at 16:16










  • As @JackD'Aurizio once did, take the log of the entire thing and differentiate.
    – Simply Beautiful Art
    Aug 13 '17 at 16:19










  • @Frank Just pointing it out, there is an alternative approached based on log differentiation.
    – Simply Beautiful Art
    Aug 13 '17 at 16:29










  • @SimplyBeautifulArt Great to know!! :D
    – Frank
    Aug 13 '17 at 16:31










  • Anyway, have a look at pages 31-32 of these notes: drive.google.com/file/d/0BxKdOVsjsuEwdjBEM1dpRkhMa2s/view
    – Jack D'Aurizio
    Aug 13 '17 at 20:04















up vote
6
down vote

favorite












You can prove the expansion$$frac {sintheta}{theta}=left{1-left(frac {theta}{pi}right)^2right}left{1-left(frac {theta}{2pi}right)^2right}ldots$$By taking the expansion$$sin nphi=2^{n-1}sinphicosphileft(sin^2frac {pi}n-sin^2phiright)left(sin^2frac {2pi}{n}-sin^2phiright)ldots$$substituting $phi=theta/n$ and dividing by$$n=2^{n-1}sinfrac {pi}nsinfrac {2pi}nldotssinfrac {pi(n-1)}{n}$$





However, when I try, I always get zero as the answer. I started off with$$sintheta=2^{n-1}sinfrac {theta}ncosfrac {theta}nleft(sin^2frac {pi}n-sin^2frac {theta}nright)left(sin^2frac {2pi}n-sin^2frac {theta}nright)ldots$$And divided it to get$$frac {sintheta}n=frac {sinfrac {theta}ncosfrac {theta}nleft(sin^2frac {pi}n-sin^2frac {theta}nright)left(sin^2frac {2pi}n-sin^2frac {theta}nright)ldots}{sinfrac {theta}ncosfrac {theta}nldotssinfrac {pi(n-1)}n}$$However, when I multiply both sides by $n$ and take the limit as $n$ tends towards infinity, I get the expansion as$$begin{align*}sintheta & =thetaleft{frac 12-frac 12left(frac {theta}nright)^2right}left{frac 13-frac 13left(frac {theta}{2pi}right)^2right}ldots\\ & =thetaprodlimits_{k=1}^{infty}left{frac 1{k+1}-frac 1{k+1}left(frac {theta}{kpi}right)^2right}\ & =0end{align*}$$




Question: I'm trying to prove$$frac {sintheta}{theta}=prodlimits_{kgeq1}left{1-left(frac {theta}{kpi}right)^2right}$$So where did I go wrong?











share|cite|improve this question
























  • Do you want to prove the pointwise convergence for any $thetainmathbb{R}$ or $inmathbb{C}$? Or the uniform convergence over compact subsets of $mathbb{R}$ or $mathbb{C}$?
    – Jack D'Aurizio
    Aug 13 '17 at 16:16










  • As @JackD'Aurizio once did, take the log of the entire thing and differentiate.
    – Simply Beautiful Art
    Aug 13 '17 at 16:19










  • @Frank Just pointing it out, there is an alternative approached based on log differentiation.
    – Simply Beautiful Art
    Aug 13 '17 at 16:29










  • @SimplyBeautifulArt Great to know!! :D
    – Frank
    Aug 13 '17 at 16:31










  • Anyway, have a look at pages 31-32 of these notes: drive.google.com/file/d/0BxKdOVsjsuEwdjBEM1dpRkhMa2s/view
    – Jack D'Aurizio
    Aug 13 '17 at 20:04













up vote
6
down vote

favorite









up vote
6
down vote

favorite











You can prove the expansion$$frac {sintheta}{theta}=left{1-left(frac {theta}{pi}right)^2right}left{1-left(frac {theta}{2pi}right)^2right}ldots$$By taking the expansion$$sin nphi=2^{n-1}sinphicosphileft(sin^2frac {pi}n-sin^2phiright)left(sin^2frac {2pi}{n}-sin^2phiright)ldots$$substituting $phi=theta/n$ and dividing by$$n=2^{n-1}sinfrac {pi}nsinfrac {2pi}nldotssinfrac {pi(n-1)}{n}$$





However, when I try, I always get zero as the answer. I started off with$$sintheta=2^{n-1}sinfrac {theta}ncosfrac {theta}nleft(sin^2frac {pi}n-sin^2frac {theta}nright)left(sin^2frac {2pi}n-sin^2frac {theta}nright)ldots$$And divided it to get$$frac {sintheta}n=frac {sinfrac {theta}ncosfrac {theta}nleft(sin^2frac {pi}n-sin^2frac {theta}nright)left(sin^2frac {2pi}n-sin^2frac {theta}nright)ldots}{sinfrac {theta}ncosfrac {theta}nldotssinfrac {pi(n-1)}n}$$However, when I multiply both sides by $n$ and take the limit as $n$ tends towards infinity, I get the expansion as$$begin{align*}sintheta & =thetaleft{frac 12-frac 12left(frac {theta}nright)^2right}left{frac 13-frac 13left(frac {theta}{2pi}right)^2right}ldots\\ & =thetaprodlimits_{k=1}^{infty}left{frac 1{k+1}-frac 1{k+1}left(frac {theta}{kpi}right)^2right}\ & =0end{align*}$$




Question: I'm trying to prove$$frac {sintheta}{theta}=prodlimits_{kgeq1}left{1-left(frac {theta}{kpi}right)^2right}$$So where did I go wrong?











share|cite|improve this question















You can prove the expansion$$frac {sintheta}{theta}=left{1-left(frac {theta}{pi}right)^2right}left{1-left(frac {theta}{2pi}right)^2right}ldots$$By taking the expansion$$sin nphi=2^{n-1}sinphicosphileft(sin^2frac {pi}n-sin^2phiright)left(sin^2frac {2pi}{n}-sin^2phiright)ldots$$substituting $phi=theta/n$ and dividing by$$n=2^{n-1}sinfrac {pi}nsinfrac {2pi}nldotssinfrac {pi(n-1)}{n}$$





However, when I try, I always get zero as the answer. I started off with$$sintheta=2^{n-1}sinfrac {theta}ncosfrac {theta}nleft(sin^2frac {pi}n-sin^2frac {theta}nright)left(sin^2frac {2pi}n-sin^2frac {theta}nright)ldots$$And divided it to get$$frac {sintheta}n=frac {sinfrac {theta}ncosfrac {theta}nleft(sin^2frac {pi}n-sin^2frac {theta}nright)left(sin^2frac {2pi}n-sin^2frac {theta}nright)ldots}{sinfrac {theta}ncosfrac {theta}nldotssinfrac {pi(n-1)}n}$$However, when I multiply both sides by $n$ and take the limit as $n$ tends towards infinity, I get the expansion as$$begin{align*}sintheta & =thetaleft{frac 12-frac 12left(frac {theta}nright)^2right}left{frac 13-frac 13left(frac {theta}{2pi}right)^2right}ldots\\ & =thetaprodlimits_{k=1}^{infty}left{frac 1{k+1}-frac 1{k+1}left(frac {theta}{kpi}right)^2right}\ & =0end{align*}$$




Question: I'm trying to prove$$frac {sintheta}{theta}=prodlimits_{kgeq1}left{1-left(frac {theta}{kpi}right)^2right}$$So where did I go wrong?








trigonometry






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edited Aug 14 '17 at 0:29









Blue

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asked Aug 13 '17 at 16:12









Frank

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  • Do you want to prove the pointwise convergence for any $thetainmathbb{R}$ or $inmathbb{C}$? Or the uniform convergence over compact subsets of $mathbb{R}$ or $mathbb{C}$?
    – Jack D'Aurizio
    Aug 13 '17 at 16:16










  • As @JackD'Aurizio once did, take the log of the entire thing and differentiate.
    – Simply Beautiful Art
    Aug 13 '17 at 16:19










  • @Frank Just pointing it out, there is an alternative approached based on log differentiation.
    – Simply Beautiful Art
    Aug 13 '17 at 16:29










  • @SimplyBeautifulArt Great to know!! :D
    – Frank
    Aug 13 '17 at 16:31










  • Anyway, have a look at pages 31-32 of these notes: drive.google.com/file/d/0BxKdOVsjsuEwdjBEM1dpRkhMa2s/view
    – Jack D'Aurizio
    Aug 13 '17 at 20:04


















  • Do you want to prove the pointwise convergence for any $thetainmathbb{R}$ or $inmathbb{C}$? Or the uniform convergence over compact subsets of $mathbb{R}$ or $mathbb{C}$?
    – Jack D'Aurizio
    Aug 13 '17 at 16:16










  • As @JackD'Aurizio once did, take the log of the entire thing and differentiate.
    – Simply Beautiful Art
    Aug 13 '17 at 16:19










  • @Frank Just pointing it out, there is an alternative approached based on log differentiation.
    – Simply Beautiful Art
    Aug 13 '17 at 16:29










  • @SimplyBeautifulArt Great to know!! :D
    – Frank
    Aug 13 '17 at 16:31










  • Anyway, have a look at pages 31-32 of these notes: drive.google.com/file/d/0BxKdOVsjsuEwdjBEM1dpRkhMa2s/view
    – Jack D'Aurizio
    Aug 13 '17 at 20:04
















Do you want to prove the pointwise convergence for any $thetainmathbb{R}$ or $inmathbb{C}$? Or the uniform convergence over compact subsets of $mathbb{R}$ or $mathbb{C}$?
– Jack D'Aurizio
Aug 13 '17 at 16:16




Do you want to prove the pointwise convergence for any $thetainmathbb{R}$ or $inmathbb{C}$? Or the uniform convergence over compact subsets of $mathbb{R}$ or $mathbb{C}$?
– Jack D'Aurizio
Aug 13 '17 at 16:16












As @JackD'Aurizio once did, take the log of the entire thing and differentiate.
– Simply Beautiful Art
Aug 13 '17 at 16:19




As @JackD'Aurizio once did, take the log of the entire thing and differentiate.
– Simply Beautiful Art
Aug 13 '17 at 16:19












@Frank Just pointing it out, there is an alternative approached based on log differentiation.
– Simply Beautiful Art
Aug 13 '17 at 16:29




@Frank Just pointing it out, there is an alternative approached based on log differentiation.
– Simply Beautiful Art
Aug 13 '17 at 16:29












@SimplyBeautifulArt Great to know!! :D
– Frank
Aug 13 '17 at 16:31




@SimplyBeautifulArt Great to know!! :D
– Frank
Aug 13 '17 at 16:31












Anyway, have a look at pages 31-32 of these notes: drive.google.com/file/d/0BxKdOVsjsuEwdjBEM1dpRkhMa2s/view
– Jack D'Aurizio
Aug 13 '17 at 20:04




Anyway, have a look at pages 31-32 of these notes: drive.google.com/file/d/0BxKdOVsjsuEwdjBEM1dpRkhMa2s/view
– Jack D'Aurizio
Aug 13 '17 at 20:04










1 Answer
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We have
begin{align}
sintheta&=2sinfrac{theta}{2}cosfrac{theta}{2}\
&=2sinfrac{theta}{2}sinleft(frac{pi}{2}+frac{theta}{2}right)..........(1).
end{align}

Similarly in (1) changing $theta$ into $frac{theta}{2}$ and $frac{pi}{2}+frac{theta}{2}$ successively, we have
$$sinfrac{theta}{2}=2sinfrac{theta}{2^2}sinleft(frac{pi}{2}+frac{theta}{2^2}right)=2sinfrac{theta}{2^2}sinleft(frac{2pi}{2^2}+frac{theta}{2^2}right),$$
and
begin{align}
sinleft(frac{pi}{2}+frac{theta}{2}right)&=2sinleft(frac{pi}{2^2}+frac{theta}{2^2}right)cdotsinleft(frac{pi}{2}+frac{pi}{2^2}+frac{theta}{2^2}right)\
&=2sinleft(frac{pi}{2^2}+frac{theta}{2^2}right)cdotsinleft(frac{3pi}{2^2}+frac{theta}{2^2}right).
end{align}

Substituting these values in the right-hand side of (1) we have, after rearranging,
$$sintheta=2^3sinfrac{theta}{2^2}sinfrac{pi+theta}{2^2}sinfrac{2pi+theta}{2^2}sinfrac{3pi+theta}{2^2}.........(2).$$
Continuing thus, we can finally have
$$sintheta=2^{p-1}sinfrac{theta}{p}sinfrac{pi+theta}{p}sinfrac{2pi+theta}{p}...sinfrac{(p-1)pi+theta}{p} ......(3),$$
where $p$ is a power of 2.



The last factor in (3)
$$=sinleft[pi-frac{pi-theta}{p}right]=sinfrac{pi-theta}{p}.$$
The last but one
$$=sinfrac{(p-2)pi+theta}{p}=sinfrac{2pi-theta}{p},$$
and so on.



Hence, taking together the second and the last factors, the third and the next to last, and so on, the equation (3) becomes
$$sintheta=2^{p-1}sinfrac{theta}{p}left{sinfrac{pi+theta}{p}sinfrac{pi-theta}{p}right}left{sinfrac{2pi+theta}{p}sinfrac{2pi-theta}{p}right}... ....(4).$$
The last factor is
$$sinfrac{frac{p}{2}pi+theta}{p}\=sinleft(frac{pi}{2}+frac{theta}{p}right)=cosfrac{theta}{p}.$$
Hence (4) is
$$sintheta=2^{p-1}sinfrac{theta}{p}left[sin^2frac{pi}{p}-sin^2frac{theta}{p}right]left[sin^2frac{2pi}{p}-sin^2frac{theta}{p}right]\...left[sin^2frac{left(frac{p}{2}-1right)pi}{p}-sin^2frac{theta}{p}right]cdotcosfrac{theta}{p}.........(5)$$
Divide both sides of (5) by $sinfrac{theta}{p}$ and let $thetato 0$. Then $lim_{thetato 0}frac{sintheta}{sinfrac{theta}{p}}=p,$ and we have
$$p=2^{p-1}cdotsin^2frac{pi}{p}cdotsin^2frac{2pi}{p}cdotsin^2frac{3pi}{p}...sin^2frac{left(frac{p}{2}-1right)pi}{p}............(6)$$
Dividing (5) by (6), we have
$$sintheta=psinfrac{theta}{p}left[1-frac{sin^2frac{theta}{p}}{sin^2frac{pi}{p}}right]left[1-frac{sin^2frac{theta}{p}}{sin^2frac{2pi}{p}}right]left[1-frac{sin^2frac{theta}{p}}{sin^2frac{3pi}{p}}right]...\...left[1-frac{sin^2frac{theta}{p}}{sin^2frac{left(frac{p}{2}-1right)pi}{p}}right]cosfrac{theta}{p}.......(7)$$
Now let $ptoinfty$.
Since
begin{align}
lim_{ptoinfty}left[psinfrac{theta}{p}right]&=theta,\
lim_{ptoinfty}left[frac{sin^2frac{theta}{p}}{sin^2frac{pi}{p}}right]&=frac{theta^2}{pi^2},
end{align}

and so on, we have
$$sintheta=thetaleft(1-frac{theta^2}{pi^2}right)left(1-frac{theta^2}{2^2pi^2}right)left(1-frac{theta^2}{3^2pi^2}right)...$$






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  • Can you explain to me how you jumped from $(4)$ to $(5)$? I'm not sure what you used to get the squared sine to show up...
    – Frank W.
    Nov 21 at 20:30






  • 1




    @Frank W. : I've used the trigonometric identity $$ sin{(A+B)} times sin{(A-B)} = sin^{2}A-sin^{2}B $$
    – Anant
    Nov 27 at 4:21












  • Okay I don't know why I didn't think to expand the sine functions. Thanks though!
    – Frank W.
    Dec 3 at 15:51











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We have
begin{align}
sintheta&=2sinfrac{theta}{2}cosfrac{theta}{2}\
&=2sinfrac{theta}{2}sinleft(frac{pi}{2}+frac{theta}{2}right)..........(1).
end{align}

Similarly in (1) changing $theta$ into $frac{theta}{2}$ and $frac{pi}{2}+frac{theta}{2}$ successively, we have
$$sinfrac{theta}{2}=2sinfrac{theta}{2^2}sinleft(frac{pi}{2}+frac{theta}{2^2}right)=2sinfrac{theta}{2^2}sinleft(frac{2pi}{2^2}+frac{theta}{2^2}right),$$
and
begin{align}
sinleft(frac{pi}{2}+frac{theta}{2}right)&=2sinleft(frac{pi}{2^2}+frac{theta}{2^2}right)cdotsinleft(frac{pi}{2}+frac{pi}{2^2}+frac{theta}{2^2}right)\
&=2sinleft(frac{pi}{2^2}+frac{theta}{2^2}right)cdotsinleft(frac{3pi}{2^2}+frac{theta}{2^2}right).
end{align}

Substituting these values in the right-hand side of (1) we have, after rearranging,
$$sintheta=2^3sinfrac{theta}{2^2}sinfrac{pi+theta}{2^2}sinfrac{2pi+theta}{2^2}sinfrac{3pi+theta}{2^2}.........(2).$$
Continuing thus, we can finally have
$$sintheta=2^{p-1}sinfrac{theta}{p}sinfrac{pi+theta}{p}sinfrac{2pi+theta}{p}...sinfrac{(p-1)pi+theta}{p} ......(3),$$
where $p$ is a power of 2.



The last factor in (3)
$$=sinleft[pi-frac{pi-theta}{p}right]=sinfrac{pi-theta}{p}.$$
The last but one
$$=sinfrac{(p-2)pi+theta}{p}=sinfrac{2pi-theta}{p},$$
and so on.



Hence, taking together the second and the last factors, the third and the next to last, and so on, the equation (3) becomes
$$sintheta=2^{p-1}sinfrac{theta}{p}left{sinfrac{pi+theta}{p}sinfrac{pi-theta}{p}right}left{sinfrac{2pi+theta}{p}sinfrac{2pi-theta}{p}right}... ....(4).$$
The last factor is
$$sinfrac{frac{p}{2}pi+theta}{p}\=sinleft(frac{pi}{2}+frac{theta}{p}right)=cosfrac{theta}{p}.$$
Hence (4) is
$$sintheta=2^{p-1}sinfrac{theta}{p}left[sin^2frac{pi}{p}-sin^2frac{theta}{p}right]left[sin^2frac{2pi}{p}-sin^2frac{theta}{p}right]\...left[sin^2frac{left(frac{p}{2}-1right)pi}{p}-sin^2frac{theta}{p}right]cdotcosfrac{theta}{p}.........(5)$$
Divide both sides of (5) by $sinfrac{theta}{p}$ and let $thetato 0$. Then $lim_{thetato 0}frac{sintheta}{sinfrac{theta}{p}}=p,$ and we have
$$p=2^{p-1}cdotsin^2frac{pi}{p}cdotsin^2frac{2pi}{p}cdotsin^2frac{3pi}{p}...sin^2frac{left(frac{p}{2}-1right)pi}{p}............(6)$$
Dividing (5) by (6), we have
$$sintheta=psinfrac{theta}{p}left[1-frac{sin^2frac{theta}{p}}{sin^2frac{pi}{p}}right]left[1-frac{sin^2frac{theta}{p}}{sin^2frac{2pi}{p}}right]left[1-frac{sin^2frac{theta}{p}}{sin^2frac{3pi}{p}}right]...\...left[1-frac{sin^2frac{theta}{p}}{sin^2frac{left(frac{p}{2}-1right)pi}{p}}right]cosfrac{theta}{p}.......(7)$$
Now let $ptoinfty$.
Since
begin{align}
lim_{ptoinfty}left[psinfrac{theta}{p}right]&=theta,\
lim_{ptoinfty}left[frac{sin^2frac{theta}{p}}{sin^2frac{pi}{p}}right]&=frac{theta^2}{pi^2},
end{align}

and so on, we have
$$sintheta=thetaleft(1-frac{theta^2}{pi^2}right)left(1-frac{theta^2}{2^2pi^2}right)left(1-frac{theta^2}{3^2pi^2}right)...$$






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  • Can you explain to me how you jumped from $(4)$ to $(5)$? I'm not sure what you used to get the squared sine to show up...
    – Frank W.
    Nov 21 at 20:30






  • 1




    @Frank W. : I've used the trigonometric identity $$ sin{(A+B)} times sin{(A-B)} = sin^{2}A-sin^{2}B $$
    – Anant
    Nov 27 at 4:21












  • Okay I don't know why I didn't think to expand the sine functions. Thanks though!
    – Frank W.
    Dec 3 at 15:51















up vote
1
down vote













We have
begin{align}
sintheta&=2sinfrac{theta}{2}cosfrac{theta}{2}\
&=2sinfrac{theta}{2}sinleft(frac{pi}{2}+frac{theta}{2}right)..........(1).
end{align}

Similarly in (1) changing $theta$ into $frac{theta}{2}$ and $frac{pi}{2}+frac{theta}{2}$ successively, we have
$$sinfrac{theta}{2}=2sinfrac{theta}{2^2}sinleft(frac{pi}{2}+frac{theta}{2^2}right)=2sinfrac{theta}{2^2}sinleft(frac{2pi}{2^2}+frac{theta}{2^2}right),$$
and
begin{align}
sinleft(frac{pi}{2}+frac{theta}{2}right)&=2sinleft(frac{pi}{2^2}+frac{theta}{2^2}right)cdotsinleft(frac{pi}{2}+frac{pi}{2^2}+frac{theta}{2^2}right)\
&=2sinleft(frac{pi}{2^2}+frac{theta}{2^2}right)cdotsinleft(frac{3pi}{2^2}+frac{theta}{2^2}right).
end{align}

Substituting these values in the right-hand side of (1) we have, after rearranging,
$$sintheta=2^3sinfrac{theta}{2^2}sinfrac{pi+theta}{2^2}sinfrac{2pi+theta}{2^2}sinfrac{3pi+theta}{2^2}.........(2).$$
Continuing thus, we can finally have
$$sintheta=2^{p-1}sinfrac{theta}{p}sinfrac{pi+theta}{p}sinfrac{2pi+theta}{p}...sinfrac{(p-1)pi+theta}{p} ......(3),$$
where $p$ is a power of 2.



The last factor in (3)
$$=sinleft[pi-frac{pi-theta}{p}right]=sinfrac{pi-theta}{p}.$$
The last but one
$$=sinfrac{(p-2)pi+theta}{p}=sinfrac{2pi-theta}{p},$$
and so on.



Hence, taking together the second and the last factors, the third and the next to last, and so on, the equation (3) becomes
$$sintheta=2^{p-1}sinfrac{theta}{p}left{sinfrac{pi+theta}{p}sinfrac{pi-theta}{p}right}left{sinfrac{2pi+theta}{p}sinfrac{2pi-theta}{p}right}... ....(4).$$
The last factor is
$$sinfrac{frac{p}{2}pi+theta}{p}\=sinleft(frac{pi}{2}+frac{theta}{p}right)=cosfrac{theta}{p}.$$
Hence (4) is
$$sintheta=2^{p-1}sinfrac{theta}{p}left[sin^2frac{pi}{p}-sin^2frac{theta}{p}right]left[sin^2frac{2pi}{p}-sin^2frac{theta}{p}right]\...left[sin^2frac{left(frac{p}{2}-1right)pi}{p}-sin^2frac{theta}{p}right]cdotcosfrac{theta}{p}.........(5)$$
Divide both sides of (5) by $sinfrac{theta}{p}$ and let $thetato 0$. Then $lim_{thetato 0}frac{sintheta}{sinfrac{theta}{p}}=p,$ and we have
$$p=2^{p-1}cdotsin^2frac{pi}{p}cdotsin^2frac{2pi}{p}cdotsin^2frac{3pi}{p}...sin^2frac{left(frac{p}{2}-1right)pi}{p}............(6)$$
Dividing (5) by (6), we have
$$sintheta=psinfrac{theta}{p}left[1-frac{sin^2frac{theta}{p}}{sin^2frac{pi}{p}}right]left[1-frac{sin^2frac{theta}{p}}{sin^2frac{2pi}{p}}right]left[1-frac{sin^2frac{theta}{p}}{sin^2frac{3pi}{p}}right]...\...left[1-frac{sin^2frac{theta}{p}}{sin^2frac{left(frac{p}{2}-1right)pi}{p}}right]cosfrac{theta}{p}.......(7)$$
Now let $ptoinfty$.
Since
begin{align}
lim_{ptoinfty}left[psinfrac{theta}{p}right]&=theta,\
lim_{ptoinfty}left[frac{sin^2frac{theta}{p}}{sin^2frac{pi}{p}}right]&=frac{theta^2}{pi^2},
end{align}

and so on, we have
$$sintheta=thetaleft(1-frac{theta^2}{pi^2}right)left(1-frac{theta^2}{2^2pi^2}right)left(1-frac{theta^2}{3^2pi^2}right)...$$






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  • Can you explain to me how you jumped from $(4)$ to $(5)$? I'm not sure what you used to get the squared sine to show up...
    – Frank W.
    Nov 21 at 20:30






  • 1




    @Frank W. : I've used the trigonometric identity $$ sin{(A+B)} times sin{(A-B)} = sin^{2}A-sin^{2}B $$
    – Anant
    Nov 27 at 4:21












  • Okay I don't know why I didn't think to expand the sine functions. Thanks though!
    – Frank W.
    Dec 3 at 15:51













up vote
1
down vote










up vote
1
down vote









We have
begin{align}
sintheta&=2sinfrac{theta}{2}cosfrac{theta}{2}\
&=2sinfrac{theta}{2}sinleft(frac{pi}{2}+frac{theta}{2}right)..........(1).
end{align}

Similarly in (1) changing $theta$ into $frac{theta}{2}$ and $frac{pi}{2}+frac{theta}{2}$ successively, we have
$$sinfrac{theta}{2}=2sinfrac{theta}{2^2}sinleft(frac{pi}{2}+frac{theta}{2^2}right)=2sinfrac{theta}{2^2}sinleft(frac{2pi}{2^2}+frac{theta}{2^2}right),$$
and
begin{align}
sinleft(frac{pi}{2}+frac{theta}{2}right)&=2sinleft(frac{pi}{2^2}+frac{theta}{2^2}right)cdotsinleft(frac{pi}{2}+frac{pi}{2^2}+frac{theta}{2^2}right)\
&=2sinleft(frac{pi}{2^2}+frac{theta}{2^2}right)cdotsinleft(frac{3pi}{2^2}+frac{theta}{2^2}right).
end{align}

Substituting these values in the right-hand side of (1) we have, after rearranging,
$$sintheta=2^3sinfrac{theta}{2^2}sinfrac{pi+theta}{2^2}sinfrac{2pi+theta}{2^2}sinfrac{3pi+theta}{2^2}.........(2).$$
Continuing thus, we can finally have
$$sintheta=2^{p-1}sinfrac{theta}{p}sinfrac{pi+theta}{p}sinfrac{2pi+theta}{p}...sinfrac{(p-1)pi+theta}{p} ......(3),$$
where $p$ is a power of 2.



The last factor in (3)
$$=sinleft[pi-frac{pi-theta}{p}right]=sinfrac{pi-theta}{p}.$$
The last but one
$$=sinfrac{(p-2)pi+theta}{p}=sinfrac{2pi-theta}{p},$$
and so on.



Hence, taking together the second and the last factors, the third and the next to last, and so on, the equation (3) becomes
$$sintheta=2^{p-1}sinfrac{theta}{p}left{sinfrac{pi+theta}{p}sinfrac{pi-theta}{p}right}left{sinfrac{2pi+theta}{p}sinfrac{2pi-theta}{p}right}... ....(4).$$
The last factor is
$$sinfrac{frac{p}{2}pi+theta}{p}\=sinleft(frac{pi}{2}+frac{theta}{p}right)=cosfrac{theta}{p}.$$
Hence (4) is
$$sintheta=2^{p-1}sinfrac{theta}{p}left[sin^2frac{pi}{p}-sin^2frac{theta}{p}right]left[sin^2frac{2pi}{p}-sin^2frac{theta}{p}right]\...left[sin^2frac{left(frac{p}{2}-1right)pi}{p}-sin^2frac{theta}{p}right]cdotcosfrac{theta}{p}.........(5)$$
Divide both sides of (5) by $sinfrac{theta}{p}$ and let $thetato 0$. Then $lim_{thetato 0}frac{sintheta}{sinfrac{theta}{p}}=p,$ and we have
$$p=2^{p-1}cdotsin^2frac{pi}{p}cdotsin^2frac{2pi}{p}cdotsin^2frac{3pi}{p}...sin^2frac{left(frac{p}{2}-1right)pi}{p}............(6)$$
Dividing (5) by (6), we have
$$sintheta=psinfrac{theta}{p}left[1-frac{sin^2frac{theta}{p}}{sin^2frac{pi}{p}}right]left[1-frac{sin^2frac{theta}{p}}{sin^2frac{2pi}{p}}right]left[1-frac{sin^2frac{theta}{p}}{sin^2frac{3pi}{p}}right]...\...left[1-frac{sin^2frac{theta}{p}}{sin^2frac{left(frac{p}{2}-1right)pi}{p}}right]cosfrac{theta}{p}.......(7)$$
Now let $ptoinfty$.
Since
begin{align}
lim_{ptoinfty}left[psinfrac{theta}{p}right]&=theta,\
lim_{ptoinfty}left[frac{sin^2frac{theta}{p}}{sin^2frac{pi}{p}}right]&=frac{theta^2}{pi^2},
end{align}

and so on, we have
$$sintheta=thetaleft(1-frac{theta^2}{pi^2}right)left(1-frac{theta^2}{2^2pi^2}right)left(1-frac{theta^2}{3^2pi^2}right)...$$






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We have
begin{align}
sintheta&=2sinfrac{theta}{2}cosfrac{theta}{2}\
&=2sinfrac{theta}{2}sinleft(frac{pi}{2}+frac{theta}{2}right)..........(1).
end{align}

Similarly in (1) changing $theta$ into $frac{theta}{2}$ and $frac{pi}{2}+frac{theta}{2}$ successively, we have
$$sinfrac{theta}{2}=2sinfrac{theta}{2^2}sinleft(frac{pi}{2}+frac{theta}{2^2}right)=2sinfrac{theta}{2^2}sinleft(frac{2pi}{2^2}+frac{theta}{2^2}right),$$
and
begin{align}
sinleft(frac{pi}{2}+frac{theta}{2}right)&=2sinleft(frac{pi}{2^2}+frac{theta}{2^2}right)cdotsinleft(frac{pi}{2}+frac{pi}{2^2}+frac{theta}{2^2}right)\
&=2sinleft(frac{pi}{2^2}+frac{theta}{2^2}right)cdotsinleft(frac{3pi}{2^2}+frac{theta}{2^2}right).
end{align}

Substituting these values in the right-hand side of (1) we have, after rearranging,
$$sintheta=2^3sinfrac{theta}{2^2}sinfrac{pi+theta}{2^2}sinfrac{2pi+theta}{2^2}sinfrac{3pi+theta}{2^2}.........(2).$$
Continuing thus, we can finally have
$$sintheta=2^{p-1}sinfrac{theta}{p}sinfrac{pi+theta}{p}sinfrac{2pi+theta}{p}...sinfrac{(p-1)pi+theta}{p} ......(3),$$
where $p$ is a power of 2.



The last factor in (3)
$$=sinleft[pi-frac{pi-theta}{p}right]=sinfrac{pi-theta}{p}.$$
The last but one
$$=sinfrac{(p-2)pi+theta}{p}=sinfrac{2pi-theta}{p},$$
and so on.



Hence, taking together the second and the last factors, the third and the next to last, and so on, the equation (3) becomes
$$sintheta=2^{p-1}sinfrac{theta}{p}left{sinfrac{pi+theta}{p}sinfrac{pi-theta}{p}right}left{sinfrac{2pi+theta}{p}sinfrac{2pi-theta}{p}right}... ....(4).$$
The last factor is
$$sinfrac{frac{p}{2}pi+theta}{p}\=sinleft(frac{pi}{2}+frac{theta}{p}right)=cosfrac{theta}{p}.$$
Hence (4) is
$$sintheta=2^{p-1}sinfrac{theta}{p}left[sin^2frac{pi}{p}-sin^2frac{theta}{p}right]left[sin^2frac{2pi}{p}-sin^2frac{theta}{p}right]\...left[sin^2frac{left(frac{p}{2}-1right)pi}{p}-sin^2frac{theta}{p}right]cdotcosfrac{theta}{p}.........(5)$$
Divide both sides of (5) by $sinfrac{theta}{p}$ and let $thetato 0$. Then $lim_{thetato 0}frac{sintheta}{sinfrac{theta}{p}}=p,$ and we have
$$p=2^{p-1}cdotsin^2frac{pi}{p}cdotsin^2frac{2pi}{p}cdotsin^2frac{3pi}{p}...sin^2frac{left(frac{p}{2}-1right)pi}{p}............(6)$$
Dividing (5) by (6), we have
$$sintheta=psinfrac{theta}{p}left[1-frac{sin^2frac{theta}{p}}{sin^2frac{pi}{p}}right]left[1-frac{sin^2frac{theta}{p}}{sin^2frac{2pi}{p}}right]left[1-frac{sin^2frac{theta}{p}}{sin^2frac{3pi}{p}}right]...\...left[1-frac{sin^2frac{theta}{p}}{sin^2frac{left(frac{p}{2}-1right)pi}{p}}right]cosfrac{theta}{p}.......(7)$$
Now let $ptoinfty$.
Since
begin{align}
lim_{ptoinfty}left[psinfrac{theta}{p}right]&=theta,\
lim_{ptoinfty}left[frac{sin^2frac{theta}{p}}{sin^2frac{pi}{p}}right]&=frac{theta^2}{pi^2},
end{align}

and so on, we have
$$sintheta=thetaleft(1-frac{theta^2}{pi^2}right)left(1-frac{theta^2}{2^2pi^2}right)left(1-frac{theta^2}{3^2pi^2}right)...$$







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share|cite|improve this answer








edited Nov 27 at 4:24

























answered Jan 4 at 17:46









Anant

147




147












  • Can you explain to me how you jumped from $(4)$ to $(5)$? I'm not sure what you used to get the squared sine to show up...
    – Frank W.
    Nov 21 at 20:30






  • 1




    @Frank W. : I've used the trigonometric identity $$ sin{(A+B)} times sin{(A-B)} = sin^{2}A-sin^{2}B $$
    – Anant
    Nov 27 at 4:21












  • Okay I don't know why I didn't think to expand the sine functions. Thanks though!
    – Frank W.
    Dec 3 at 15:51


















  • Can you explain to me how you jumped from $(4)$ to $(5)$? I'm not sure what you used to get the squared sine to show up...
    – Frank W.
    Nov 21 at 20:30






  • 1




    @Frank W. : I've used the trigonometric identity $$ sin{(A+B)} times sin{(A-B)} = sin^{2}A-sin^{2}B $$
    – Anant
    Nov 27 at 4:21












  • Okay I don't know why I didn't think to expand the sine functions. Thanks though!
    – Frank W.
    Dec 3 at 15:51
















Can you explain to me how you jumped from $(4)$ to $(5)$? I'm not sure what you used to get the squared sine to show up...
– Frank W.
Nov 21 at 20:30




Can you explain to me how you jumped from $(4)$ to $(5)$? I'm not sure what you used to get the squared sine to show up...
– Frank W.
Nov 21 at 20:30




1




1




@Frank W. : I've used the trigonometric identity $$ sin{(A+B)} times sin{(A-B)} = sin^{2}A-sin^{2}B $$
– Anant
Nov 27 at 4:21






@Frank W. : I've used the trigonometric identity $$ sin{(A+B)} times sin{(A-B)} = sin^{2}A-sin^{2}B $$
– Anant
Nov 27 at 4:21














Okay I don't know why I didn't think to expand the sine functions. Thanks though!
– Frank W.
Dec 3 at 15:51




Okay I don't know why I didn't think to expand the sine functions. Thanks though!
– Frank W.
Dec 3 at 15:51


















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