Way to prove $\sin 3^{\circ}$ value I was looking wikipedia ,suddenly I saw this 
$$\sin {{3}^{{}^\circ }}=\cos {{87}^{{}^\circ }}=\frac{2\left( 1-\sqrt{3} \right)\sqrt{5+\sqrt{5}}+\left( 1+\sqrt{3} \right)\left( \sqrt{10}-\sqrt{2} \right)}{16}$$
can someone help me that : how to find $\sin3^{\circ}$ or how to prove it .
Is there a Algebraic expressions for this kind of problem ?
Thanks in advanced.
 A: First
$$\sin\frac{\pi}{5}=\sin\frac{4\pi}5=2\sin\frac{2\pi}{5}\cos\frac{2\pi}{5}=4\sin\frac{\pi}{5}\cos\frac{\pi}{5}\cos\frac{2\pi}{5}$$
Hence
$$\cos\frac{\pi}{5}\cos\frac{2\pi}{5}=\frac14$$
And
$$\cos\frac{\pi}{5}\cos\frac{2\pi}{5}=\frac{1}{2}\left(\cos\frac{\pi}{5}+\cos\frac{3\pi}{5}\right)=\frac{1}{4}$$
So, you have the system
$$\left\{\begin{eqnarray}\cos\frac{\pi}{5}\cos\frac{3\pi}{5}&=&-\frac14\\
\cos\frac{\pi}{5}+\cos\frac{3\pi}{5}&=&\frac12\end{eqnarray}\right.$$
Hence $\cos\frac{\pi}5$ and $\cos\frac{3\pi}5$ are the roots of $x^2-\frac12x-\frac14$. For this trinomial, $\Delta=\frac54$ and
$$x=\frac{1\pm\sqrt{5}}{4}$$
Since $\cos\frac{\pi}5>0$ and $\cos\frac{3\pi}5<0$, this means
$$\cos\frac{\pi}5=\frac{1+\sqrt{5}}{4}$$
And
$$\cos\frac{2\pi}{5}=-\cos\frac{3\pi}{5}=\frac{\sqrt{5}-1}{4}$$
Then
$$\sin\frac{\pi}{10}=\sin\left(\frac{\pi}{2}-\frac{2\pi}{5}\right)=\cos\frac{2\pi}{5}=\frac{\sqrt{5}-1}{4}$$
And
$$\cos\frac{\pi}{10}=\sqrt{\frac{1+\cos\frac{\pi}{5}}{2}}=\sqrt{\frac{5+\sqrt5}{8}}$$

Next
$$\cos\frac{\pi}{12}=\cos\left(\frac{\pi}{4}-\frac\pi6\right)=\frac{\sqrt2}{2}\frac{\sqrt3}{2}+\frac12\frac{\sqrt2}{2}=\frac{\sqrt2}{4}(\sqrt{3}+1)$$
$$\sin\frac{\pi}{12}=\sin\left(\frac{\pi}{4}-\frac\pi6\right)=\frac{\sqrt2}{2}\frac{\sqrt3}{2}-\frac12\frac{\sqrt2}{2}=\frac{\sqrt2}{4}(\sqrt{3}-1)$$

Finally
$$\sin 3^\circ=\sin\frac{\pi}{60}=\sin\left(\frac{\pi}{10}-\frac\pi{12}\right)\\=\left(\frac{\sqrt{5}-1}{4}\right)\left(\frac{\sqrt2}{4}(\sqrt{3}+1)\right)-\left(\sqrt{\frac{5+\sqrt5}{8}}\right)\left(\frac{\sqrt2}{4}(\sqrt{3}-1)\right)\\
=\frac{2(1-\sqrt3)\sqrt{5+\sqrt{5}}+(1+\sqrt{3})(\sqrt{10}-\sqrt{2})}{16}$$

Computability of $\cos\frac{\pi}{n}$ and $\sin\frac{\pi}{n}$ by radicals is related to constructible polygons. A more difficult result arises for $n=17$, see for instance this.
A: First it's more usual to compute this kind of formula by considering the angle in radians. You can note that :
$$3^{\circ}=\frac{\pi}{60}\text{ }\text{rad}=\left[\frac{\pi}{10}-\frac{\pi}{12}\right]\text{rad}$$
Then you can compute the sinus of $\frac{\pi}{12}$ using the formula for the triple angle on the sinus of $\frac{\pi}{4}$. For $\frac{\pi}{10}$ they present a method with a pentagon but you can also try to use the Chebyshev polynomials ...
Take a look at this article.
