Construct a polynomial equation of the least degree with rational coefficient one of root is $\sin 10^\circ$

Question

I want to construct an equation with least degree which root is $\sin 10°$, but I had no idea to find it please add a tip hint to find the equation. I want to find the solution without using the value of $\sin 10°$.

Answer 1

Hint: $\sin(3x)=3\sin x -4\sin^3 x$.

Solution:

For $x=10^\circ$, we get $\frac12 = 3\sin(10^\circ)-4\sin^3(10^\circ)$ and so $\sin(10^\circ)$ is a root $6x-8x^3=1$. By the rational root theorem, this equation has no rational roots. Therefore, $\sin(10^\circ)$ is not rational. Since $3$ is prime, this equation is the one with least degree having $\sin(10^\circ)$ as root.

Answer 2

I'm assuming degrees were intended.

you are asking about $\cos 80^\circ$ or $\cos \frac{4 \pi}{9}.$

Write $\omega = e^{2 \pi i / 9}.$ Then $$ \omega^6 + \omega^3 + 1 = 0 $$ from which we can calculate that $$ \omega + \frac{1}{\omega} \; , \; \; \omega^2 + \frac{1}{\omega^2} \; , \; \; \omega^4 + \frac{1}{\omega^4} \; , \; \; $$ are the roots of $$ x^3 - 3x + 1 \; . $$ So $2 \cos 80^\circ$ is such a root. You want half, so let $x=2t,$ which solves $$ 8 t^3 - 6 t + 1 $$

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Some detail: let us take $\beta = \omega^2 \; .$ Then $\beta^6 + \beta^3 + 1 = \omega^{12} + \omega^6 + 1 = \omega^3 + \omega^6 + 1 = 0.$

Next, $$ \left(\beta + \frac{1}{\beta}\right)^3 = \beta^3 + 3 \beta + 3 \frac{1}{\beta} + \frac{1}{\beta^3} $$ so that $$ \left(\beta + \frac{1}{\beta}\right)^3 - 3\left(\beta + \frac{1}{\beta}\right) + 1 = \beta^3 + 1 + \frac{1}{\beta^3} = \frac{\beta^6 + \beta^3 + 1}{\beta^3} = 0. $$