Find the least degree Polynomial with Integer Coefficients whose one of the roots is $ \cos(12^{\circ})$
My Try: we know that $$\cos(5x)=\cos^5x-10\cos^3x\sin^2x+5\cos x\sin^4x$$ Putting $x=12^{\circ}$ and Converting $\sin$ to $\cos$ we have
$$\frac{1}{2}=x^5-10x^3(1-x^2)+5x(1-2x^2+x^4)$$ $\implies$
$$32x^5-40x^3+10x-1=0$$
Is this the Least degree? Please let me know.
Looking at function $$f(x)=32x^5-40x^3+10x-1$$ by inspection $x=\frac 12$ is a root. So $$f(x)=(2x-1)(16 x^4+8 x^3-16 x^2-8 x+1)$$ Now, you have a quartic which can be solved with radicals and integers.
Without any quartic solving:
Set $y=2x$. The polynomial becomes $$y^5-5y^3+5y-1=(y-1)(y^4+y^3-4y^2-4y+1)$$ As $2\cos\dfrac \pi{15}\neq 1$, it is a root of $\,f(y)=y^4+y^3-4y^2-4y+1$.
Let's show this polynomial is irreducible over $\mathbf Z$. Indeed, it has no integer root (the only possibilities are $1$ and $-1$, none of which is a root). If it were reducible, it would be as the product of two (irreducible) quadratic polynomials, hence it would also be reducible modulo $2$.
However, modulo $2$, we have: $$f(y)=y^4+y^3+1.$$ It has no root in $\mathbf F_2$. Suppose we can write: $$y^4+y^3+1=(y^2+ay+b)(y^2+a'y+b')$$ By identification, we obtain at once $bb'=1$, whence $b=b'=1$ and the system: $$a+a'=1,\quad aa'=0,\quad a+a'=0,$$ which has no solution.
Thus the polynomial $y^4+y^3-4y^2-4y+1$ is irreducible, and the minimal polynomial of $\cos\dfrac \pi{15}$ is: $$16x^4+8x^3-16x^2-8x+1.$$
Hint: You made use of $\cos(5\alpha)=\frac12$ for $\alpha=12^\circ$. What is $\cos(5\alpha)$ for $\alpha=60^\circ$? So what "superfluous" root can we expect your polynomial to have?
