How to find the polynomial such that ... Let $P(x)$ be the polynomial of degree 4 and $\sin\dfrac{\pi}{24}$, $\sin\dfrac{7\pi}{24}$, $\sin\dfrac{13\pi}{24}$, $\sin\dfrac{19\pi}{24}$ are roots of $P(x)$ .
How to find $P(x)$?
Thank you very much.

Thank you every one.
But consider this problem.
Find the polynomial with degree 3 such that $\cos\dfrac{\pi}{12}$, $\cos\dfrac{9\pi}{12}$, $\cos\dfrac{17\pi}{12}$ are roots
Note that $\dfrac{\pi}{12}$, $\dfrac{9\pi}{12}$, $\dfrac{17\pi}{12}$ are solution of equation $\cos3\theta=\dfrac{1}{\sqrt{2}}$ and  $\cos\dfrac{\pi}{12}$, $\cos\dfrac{9\pi}{12}$, $\cos\dfrac{17\pi}{12}$ are distinct number.
We have $\cos3\theta=4\cos^3\theta-3\cos\theta$. Let $x=\cos\theta$, therefore 
$\cos\dfrac{\pi}{12}$, $\cos\dfrac{9\pi}{12}$, $\cos\dfrac{17\pi}{12}$ are roots of $4x^3-3x=\dfrac{1}{\sqrt{2}}.$
I want method similar to this to find $P(x)$.
Thank you.

May be not the answer you wanted given that it is of degree 8. But it has integer coefficients, so may be of interest.
If $R_n(x)=T_n(\sqrt{1-x^2})$, where $T_n$ is the Chebyshev polynomial of degree $n$, then
$$
T_n(\sin t)=\cos nt
$$
for all $t$. Because $\cos \alpha=0$, iff $\alpha$ is an odd multiple of $\pi/2$, the 12 zeros of 
$$
R_{12}(x)=1 - 72 x^2 + 840 x^4 - 3584 x^6 + 6912 x^8 - 6144 x^{10} + 2048 x^{12}
$$
are the numbers $\sin((2j+1)\pi/24), j=0,1,2,\ldots,23$. Each zero occurs here twice, because $\sin x=\sin (\pi-x)$. We can throw away the zeros that 
correspond to $3\mid 2j+1$, for those are also zeros of 
$$
R_4(x)=1-8x^2+8x^4.
$$
This leaves us with 
$$
P(x)=\frac{R_{12}(x)}{R_4(x)}=1-64x^2+320x^4-512x^6+256x^8.
$$
In addition to the prescribed zeros $P$ vanishes at the negatives of those sines. Observe that $\sin(5\pi/24)=\sin(19\pi/24)$ et cetera.
