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find the numeral value of this
Sin(1)*Sin(2)*Sin(3)*...*Sin(89)
Note:the numbers in brackets are degree
Sin(1)*Sin(2)*Sin(3)*...*Sin(89)
Note:the numbers in brackets are degree
Yes, the next step could beIt'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
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?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?
Ok thanks morphism, somehow I missed that link before - very interesting result.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 let me get it right this time.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.