closed form expression for $\sin 10^o$? As we know that $\sin 15^o$, $\sin 30^o$,$\sin 45^o$ have simple closed form expressions as these are multiples of 3, but i have never seen any simple closed form expression for $\sin 10^o$ or simply sine for any non-multiple of 3, if there exists a closed form expression, do  help me,  


  PS. I know $\sin 10^o$ is solution of $8x^3-6x+1=0$ but i can't solve it as its too tedious.  


Why is being multiple of 3 such a great thing for an angle???
Being a multiple of $3°$ isn't a great thing, as degrees are an arbitrary unit. It matters much more to be a small fraction of a full turn, as this leads to polynomial equations of a low degree.

E.g.

$$\sin 3x=-4\sin^3x+3\sin x=0$$ leads to the well-known $$\sin\frac\pi3=\frac{\sqrt3}2.$$

Some other fractions lead to closed-form expressions, but most others not, as explained by the Abel-Ruffini theorem. In particular, angle trisection (dividing by three) involves a cubic equation which is in general not solvable (the above case being an exception).

It is interesting to note that there are analytical formulas to solve cubics, but for some values of the coefficients (the so-called casus irreductibilis), the solution requires… trigonometric functions, with angle trisection, and you are circling in rounds.
I don't think that it's tedious.

Let $x=\cos\alpha.$

Thus, we need to solve $$4\cos^3\alpha-3\cos\alpha=-\frac{1}{2}$$ or$$\cos3\alpha=-\frac{1}{2},$$which gives $$3\alpha=120^{\circ}+360^{\circ}k,$$ where $k\in\mathbb Z$ or
$$3\alpha=-120^{\circ}+360^{\circ}k,$$ which for $k=0$ in the first sequence and for $k=1$ and $k=2$ in the second sequence gives:
$$x_1=\cos40^{\circ},$$
$$x_2=\cos80^{\circ}=\sin10^{\circ}$$ and
$$x_3=\cos200^{\circ}=-\cos20^{\circ}.$$