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How we can find exact value of $\sin(10^\circ)$?

I tried trigonometric ways but I get this equation: $\ 8y^3-6y+1=0$ and $y = \sin(10^\circ)$ and all the roots are complex.

I saw the pages in site, but I can't find the solution. Thanks

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    $\begingroup$ en.wikipedia.org/wiki/Cubic_function#Cardano.27s_method $\endgroup$ – gamma Mar 28 '16 at 23:56
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    $\begingroup$ Using the formula for $\sin(3\theta)$, we can get a cubic. That is likely how you got yours. But we end up with the casus irreducibilis case of the cubic, and there is no useful "algebraic" way to solve the equation. $\endgroup$ – André Nicolas Mar 29 '16 at 0:02
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    $\begingroup$ To be more specific, there is no expression for $\sin(10^\circ)$ that uses the ordinary arithmetical operations and $n$-th roots ($n=2,3,4,5,\dots$) of real numbers. $\endgroup$ – André Nicolas Mar 29 '16 at 0:11
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    $\begingroup$ S.H.W. WA (or any other CAS) outputs the roots using cubic roots of complex numbers. But to write down those cubic roots requires you to know the values of trig functions of some angle $\theta/3$, where $\theta$ is in some sense given. Basically it means that you are running in circles. There is a formula for $\sin 10^\circ$ in terms of cubic roots of complex numbers, but to calculate those cubic roots you need to know $\sin 10^\circ$. I guess this is more or less what André was telling. $\endgroup$ – Jyrki Lahtonen Mar 29 '16 at 7:35
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    $\begingroup$ It looks like there are three real roots. $\endgroup$ – robjohn Mar 29 '16 at 14:35
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Using Precise-Rewritten method for Sin 10∘, Exact value may be determined as follows:

+45..............45.....................+2

-22.5............22.5...................-2

-11.25............11.25................+2

-5.625............5.625................+2

+2.8125............8.4375............-2

+1.40625...........9.84375...........+2

+0.703125............10.546875........+2

-0.3515625..........10.1953125......-2

Hint for table: First column is half of earlier (called as Central) with sign towards target angle 10∘; second is cumulative of the first column; and the third is division of current Central with earlier Central.

If we follow above pattern infinitely, the sum will be 10. However, we can see the repeating pattern in third column (-2, +2, +2); hence this is indication for the exact value.

Now write the third column top-to-down approach as:$$+2\overline{-2+2+2} $$.

In the next step just replace 2 by √(2; the result will be : $$√(2 \overline{-√(2+√(2+√(2} $$.

Sin 10∘ will be half of this (closing brackets are collapsed by "]"). Therefore, Sin 10∘ = $$√(2 \overline{-√(2+√(2+√(2}]/2 $$

Above method is called as Precise-Rewritten method. You can find exact trigonometric value of all integer angles using this method.

I apologize above bad formatting, discourage to explain the details here.

Source: Breaking Classical Rules in Trigonometry:Precise-Rewritten method, 2016

Request to the scholars: You may assist to copy edit the main document of above method. I have not idea to formatting and use of mathematical tools at all. I cannot format as online forum requires formatting the mathematics. If someone assisted me for copy edit on Precise-Rewritten method (and other new methods), every scholar may know the new idea for new method for exact trigonometric values. My (Breaking Classical Rules in Trigonometry- Mission 2050) un-skill on mathematics formating discouraging me to expose all of those new idea.

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  • $\begingroup$ I read your article (scribd.com/document/327973537/…) , and it doesn't make sense. $\endgroup$ – S.H.W Nov 5 '16 at 22:43
  • $\begingroup$ I try to understand your solution but it was too hard to understand . and I found several problems in your article. Thank you $\endgroup$ – S.H.W Nov 5 '16 at 22:47
  • $\begingroup$ Of course this is hard> There are two reasons: $\endgroup$ – Bhava Nath Dahal Nov 6 '16 at 10:17
  • $\begingroup$ Of course, this is hard to understand. The reasons are two: first my skill to writing is awkward and second this is new method than classical system. Request to improve the language in the answer above. $\endgroup$ – Bhava Nath Dahal Nov 6 '16 at 10:21
  • $\begingroup$ Yes , that's right. $\endgroup$ – S.H.W Nov 6 '16 at 11:45

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