Numeral value of this Sin(1)*Sin(2)*Sin(3)* *Sin(89)

Is this posted as a challenge for others, or do you want to try to solve it for yourself with some help?

 

haha, this is a pretty cool little problem.

 

Pair products of sines whose degrees add up to 90 i.e.
Sin(1)*Sin(2)*Sin(3)*...*Sin(89)=sin(1)*sin(89) * sin(2)*sin(88) * .... * sin(44)*sin(46) *sin(45)

Use the trigonometric product to sum formula, and cos(90)=0 to obtain

2^(-44) * cos(88)*cos(86)*cos(84)* ... *cos(2) * sin(45) = 2^(-44.5) * cos(88)*cos(2) * cos(86)* cos(4) * ... cos(46)*cos(44)

I guess that successive application of the product to sum forumula can now be used, but am too lazy to investigate further.

Numerical calculation gives an answer of about ~ 2^(-85.75) if it's any help.

 

It's not quite so simple as successive applications of the exact same trick. If you went one further, then you would get pairings that don't quite add up to 90.

86 and 2; 82 and 6; 78 and 10, etc all add to 88

 

Yes, the next step could be

2^(-66.5) * sin(4) * sin(8) * sin(12) ... * sin(84) * sin(88)


And then....?

 

But the answer is 0?

 

The answer is 0 if and only if one of the terms of the product is zero. sin is not zero for any of the angles 1,2,3,..,89 degrees.

 

It is very helpful if you know this particular result:

http://mathforum.org/library/drmath/view/65389.html

 

For some reason I swear there is a nifty way to simplify [tex] \left( x - \frac{1}{x} \right)\left( x^2 - \frac{1}{x^2} \right)\left( x^3 - \frac{1}{x^3} \right).....\left( x^{89} - \frac{1}{x^{89}} \right)[/tex] but I can't see it right now. If so, then this problem is easily solved.

 

That would be completely ignoring the possibly useful fact that cos(90) = 0. It is possible there is a nifty simplification for that product, but it would provide a much more general solution than needed.

 

I got the same 22 term simplification as Kittel Knight. Has anyone found a better solution? Does someone want to post a clue as to the form of the answer we can expect?

 

Look at the link yenchin posted. Another way to obtain the result in there is to consider the roots of the equation x^n-1=0, i.e. the nth roots of unity given by 1, z, z^2, ..., z^(n-1), where z=e^(2pi*i/n). This leads us to the equation 1+x+...+x^(n-1)=(x-z)(x-z^2)...(x-z^(n-1)). Plug in x=1 and take absolute values. Finally, try to find a nice expression for |1-z^k|. What do you get?

 

Ok thanks morphism, somehow I missed that link before - very interesting result.

Anyway it follows immediately from that result that the sine product posed in this thread can be simplified to [itex]\sqrt{179/2} \, \times \, 2^{-89} [/itex]

Thanks.

 

Ok let me get it right this time.

[itex]\sqrt{90} \, \times \, 2^{-89} [/itex]