find exact value of $\sin(10^\circ)$

Question

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

Answer 1

Using De Moivre's Theorem to make a (circular logic) triple angle theorem, $$ \sin\left(\frac{\theta}{3}\right) = \frac{(\cos\theta+i\sin\theta)^{\frac{1}{3}} - (\cos\theta-i\sin\theta)^{\frac{1}{3}}}{2i} $$

then,

$$ \sin{10^\circ} = \frac{\sqrt[3]{\frac{\sqrt{3} + i}{2}} - \sqrt[3]{\frac{\sqrt{3} - i}{2}}}{2i}, $$ which you can simplify further to $$ = \frac{\sqrt[\leftroot{0}\uproot{3}\scriptstyle 3]{4\sqrt{3}+4i} - \sqrt[\leftroot{0}\uproot{3}\scriptstyle 3]{4\sqrt{3}-4i}}{4i} $$

Although this solution is a little redundant as it has a complex number under a cube root, it does obtain a "solution"

Answer 2

According to the comments on this question, let's solve it by taking the help of the comments.

We know that $$\sin(3x)=3\sin(x)-4\sin^{3}(x)$$

Now let's put $x=10°$

$\implies \sin(30°)=3\sin(10°)-4\sin^{3}(10°)$

Now we again all know that $\sin(30°)=\frac{1}{2}$

$\implies \frac{1}{2}=3y-4y^{3}$

Where $y=\sin(10°)$

$\implies 6y-8y^{3}=1$

$\implies 8y^{3}-6y+1=0$

$\implies y^{3}-\frac{3}{4}y+\frac{1}{8}=0$.....$(1)$

Now I will solve this cubic by Cardan's method.

Let's assume $y=(u+v)$

$\implies y^{3}=(u^{3}+v^{3})+3uv(u+v)$

$\implies y^{3}-3(uv)y-(u^{3}+v^{3})=0$.....$(2)$

Comparing the equations $(1)$ and $(2)$ we get :

$3uv=\frac{3}{4}$

$u^{3}+v^{3}=-\frac{1}{8}$

Which again implies that :

$uv=\frac{1}{4}$

$\implies u^{3}v^{3}=\frac{1}{64}$

Now $u^{3},v^{3}$ are the roots of the equation $$t^{2}+\frac{1}{8}t+\frac{1}{64}=0$$

$\implies 64t^{2}+8t+1=0$

$\implies t=\frac{-1-\sqrt{3}i}{16},\frac{-1+\sqrt{3}i}{16}$

Where $i=\sqrt{-1}$

Which finally implies

$u^{3}=\frac{-1+\sqrt{3}i}{16}$

$\implies u=\sqrt[3]{\frac{-1+\sqrt{3}i}{16}}$

And

$v^{3}=\frac{-1-\sqrt{3}i}{16}$

$\implies v=\sqrt[3]{\frac{-1-\sqrt{3}i}{16}}$

Now Wolfram Alpha gives the values of $u$ and $v$ to be

$u=0.38302+0.32139i$

And

$v=0.38302-0.32139i$

Answer 3

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.