- #1

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Sin(1)*Sin(2)*Sin(3)*...*Sin(89)

Note:the numbers in brackets are degree

- Thread starter hadi amiri 4
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- #1

- 97

- 1

Sin(1)*Sin(2)*Sin(3)*...*Sin(89)

Note:the numbers in brackets are degree

- #2

matt grime

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- #3

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challenge of course!!!!!!!!!!!!!!!!!

- #4

- 655

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haha, this is a pretty cool little problem.

- #5

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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.

- #6

- 655

- 3

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

- #7

- 99

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Yes, the next step could be

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

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

And then....?

- #8

- 97

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this problem does not need a calculator

just some formulas and a creative brain

just some formulas and a creative brain

- #9

- 54

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But the answer is 0?

- #10

matt grime

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- #11

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

- #12

Gib Z

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- #13

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- #14

uart

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- #15

morphism

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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?

- #16

uart

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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?

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.

Last edited:

- #17

uart

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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.

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

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