Proof that $\sin10^\circ$ is irrational Today I was thinking about proving this statement, but I really could not come up with an idea at all. I want to prove that $\sin10^\circ$ is irrational. Any ideas?
 A: HINT
$\sin(3x)=\sin(2x+x)$
use this and expand as a compound angle formula on right hand side and get a polynomial on RHS in terms of $\sin(x)$ for $\sin(3x) $
you will get $\sin(3 x)=3 \sin(x)-4 \sin^3(x)$
then sub $x=10^\circ$ and solve the equation as polynomial you will get answer for $\sin(10^\circ)$ in irrational form.
A: Suppose that $x=\sin(10^\circ)$ and we want to prove the irrationality of $x$.  Then we can use the Triple Angle Formula for $\sin$ to get $-4x^3+3x = \sin(30^\circ)=\frac12$; in other words, $x$ is a solution of the equation $-8x^3+6x-1=0$.
But now that we have this equation we can use another tool, the Rational Root Theorem : any rational root $\frac pq$ of the equation must have its numerator $p$ dividing $1$, and its denominator $q$ dividing $-8$.  This implies that any rational root of the polynomial must be one of $\{\pm 1, \pm\frac12, \pm\frac14, \pm\frac18\}$; now you can test each of these values directly by plugging them in to the cubic to show that none of them is a root.
Alternately, if you don't want to go to that much trouble, we can use yet another tool: Eisenstein's Criterion.  First, we 'flip' our equation by substituting $y=\frac1x$; we know that $x$ is finite and non-zero, so $y$ is also finite (and rational, if $x$ is) and non-zero, and we can multiply by it: rewriting our equation in terms of $y$ we get $-8\frac1{y^3}+6\frac1y-1=0$, and multiplying this by $y^3$ we get $-y^3+6y^2-8=0$.  Now, Eisenstein's criteria doesn't directly apply here (because the only prime that divides our constant coefficient $8$ is $2$, but we do have $2^2=4$ dividing $8$), but we can start playing around with simple substitutions like $z=y\pm 1$, $z=y\pm2$.  Trying $z=y-1$ (so $y=z+1$) first, we discover that the equation converts to $-z^3+3z^2+9z-3=0$.  And now Eisenstein's Criterion does apply, with $p=3$, and we can conclude that this polynomial in $z$ (and so our polynomial in $y$, and so our polynomial in $x$) is irreducible over the rationals.
Incidentally, the fact that this particular polynomial ($-8x^3+6x-1$) is irreducible has consequences for a famous classical problem:
Define the degree of an algebraic number as the order of the minimal (i.e. lowest-order) polynomial that it's a zero of; this is one of many equivalent definitions (though the equivalence is a deep theorem in its own right).  Now, since $-8x^3+6x-1$ is irreducible, its roots (and in particular, $\sin(10^\circ)$) must have degree $3$; if their degree were lower, then their minimal polynomial would be a factor of $-8x^3+6x-1$.  But it's known that any number that's constructible with ruler and compass must have degree $2^n$ for some $n$; informally, compasses can take square roots but not cube roots.  Since $\sin(10^\circ)$ has degree $3$, this implies that it's not constructible with ruler and compass — and in particular, that a $10^\circ$ angle isn't constructible.  But we know that a $30^\circ$ angle is constructible, so it must be impossible to get from a $30^\circ$ angle to a $10^\circ$ angle.  In other words, trisecting arbitrary angles is impossible with ruler and compass!
