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Topic: sin 1 (Read 611 times) |
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cool_joh
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Find the exact value of sin 1.
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cool_joh
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I mean sin 10
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Aryabhatta
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Re: sin 1
« Reply #3 on: Dec 26th, 2007, 12:09pm » |
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We can calculate sin 18 (see thread in easy) and sin 15 (and cos 15).
From this we get sin 3. From that we can calculate sin 1 as it is a root of a cubic for the triple angle formula for sin.
Please don't ask for the exact value 
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ThudnBlunder
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Re: sin 1
« Reply #4 on: Dec 26th, 2007, 12:38pm » |
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on Dec 26th, 2007, 12:09pm, Aryabhatta wrote:| Please don't ask for the exact value |
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sin21o = 1 - cos21o = 1 - Z, say,
and since
cos6x = 32cos6x - 48cos4x + 18cos2x - 1
cos6o = 32Z3 - 48Z2 + 18Z - 1
and we need to solve
32Z3 - 48Z2 + 18Z = a + 1
where
a = (1/4) {7 + 5 + [6(5 + 5)]}
Mathematica gives Z = 1/2 + (1/4)(y + 1/y)
where
y = [a + (a2 - 1)]1/3 
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| « Last Edit: Dec 26th, 2007, 7:58pm by ThudnBlunder » |
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cool_joh
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I think that makes sin 1 imaginary number, because 1-Z is negative since y+1/y>=2 for positive y.
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towr
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Re: sin 1
« Reply #6 on: Dec 27th, 2007, 1:45am » |
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on Dec 26th, 2007, 9:09pm, cool_joh wrote:| I think that makes sin 1 imaginary number, because 1-Z is negative since y+1/y>=2 for positive y. |
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It also depends on whether y isn't already a complex number. We have a < 1 so a2-1 < 0 so y is complex
Obviously, sin 1 must be a real number, you can measure it.
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| « Last Edit: Dec 27th, 2007, 1:46am by towr » |
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Wikipedia, Google, Mathworld, Integer sequence DB
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SMQ
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Re: sin 1
« Reply #7 on: Dec 27th, 2007, 8:46am » |
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on Dec 27th, 2007, 1:45am, towr wrote:| We have a < 1 so a2-1 < 0 so y is complex |
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Exactly. Note, also, that |a + (a2 - 1)| = (a2 + 1 - a2) = 1, which means |y| = 1 as well, and so 1/y = y*  y + 1/y = y + y* = 2 (y): a positive real number slightly less than 2 (if we take the branch of the cube root with positive real part).
--SMQ
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| « Last Edit: Dec 27th, 2007, 8:47am by SMQ » |
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ThudnBlunder
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Re: sin 1
« Reply #8 on: Dec 30th, 2007, 5:39pm » |
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on Dec 26th, 2007, 9:09pm, cool_joh wrote:| I think that makes sin 1 imaginary number, because 1-Z is negative since y+1/y>=2 for positive y. |
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This is called Casus Irreducibilis. The problem occurs even with sin( /9), a multiple of sin(/180). It was this (and not the quadratic formula) which historically first led to the study of complex numbers.
Alternatively, if we don't limit ourselves to a finite number of roots :
sin(/180) = sin(/18 - /20) = sin(/18)*cos(/20) - sin(/20)*cos(/18)
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| « Last Edit: Jan 8th, 2008, 3:37pm by ThudnBlunder » |
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