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.
 A: 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.
A: By a tedious expansion of $P(x)=(x-r_1)(x-r_2)\ldots$ that other answers have covered or by using Vieta's formulas, you can find that
$$P(x)=x^4+\left[-\sin\left(\frac{\pi}{24}\right)-\sin\left(\frac{7\pi}{24}\right)-\sin\left(\frac{11\pi}{24}\right)-\sin\left(\frac{13\pi}{24}\right)\right]x^3$$
$$\ldots+\left[\sin\left(\frac{\pi}{24}\right)\sin\left(\frac{7\pi}{24}\right)+\sin\left(\frac{\pi}{24}\right)\sin\left(\frac{13\pi}{24}\right)+\sin\left(\frac{7\pi}{24}\right)\sin\left(\frac{13\pi}{24}\right)\right.$$
$$\left.\ldots+\sin\left(\frac{\pi}{24}\right)\sin\left(\frac{19\pi}{24}\right)+\sin\left(\frac{7\pi}{24}\right)\sin\left(\frac{19\pi}{24}\right)+\sin\left(\frac{13\pi}{24}\right)\sin\left(\frac{19\pi}{24}\right)\right]x^2$$
$$\ldots+\left[-\sin\left(\frac{\pi}{24}\right)\sin\left(\frac{7\pi}{24}\right)\sin\left(\frac{13\pi}{24}\right)-\sin\left(\frac{\pi}{24}\right)\sin\left(\frac{7\pi}{24}\right)\sin\left(\frac{19\pi}{24}\right)\right.$$
$$\left.\ldots-\sin\left(\frac{\pi}{24}\right)\sin\left(\frac{13\pi}{24}\right)\sin\left(\frac{19\pi}{24}\right)-\sin\left(\frac{7\pi}{24}\right)\sin\left(\frac{13\pi}{24}\right)\sin\left(\frac{19\pi}{24}\right)\right]x$$
$$\ldots+\left[\sin\left(\frac{\pi}{24}\right)\sin\left(\frac{7\pi}{24}\right)\sin\left(\frac{13\pi}{24}\right)\sin\left(\frac{19\pi}{24}\right)\right]$$
Using Wolfram Alpha, we can find that $\sin\left(\frac{\pi}{24}\right)=\frac{1}{2}\sqrt{2-\sqrt{2-\sqrt{3}}}$, $\sin\left(\frac{7\pi}{24}\right)=\frac{1}{2}\sqrt{2+\sqrt{2-\sqrt{3}}}$, $\sin\left(\frac{13\pi}{24}\right)=\frac{1}{2}\sqrt{2+\sqrt{2+\sqrt{3}}}$ & $\sin\left(\frac{19\pi}{24}\right)=\frac{1}{2}\sqrt{2-\sqrt{2-\sqrt{3}}}$. You could then substitute these in (I'd be interested to see what it simplifies to). Please note that with an expression so long, I'm bound to have made a mistake somewhere. Thanks for the interesting question.
Edit: Using $\sin(\ldots)$ in terms of $e^{(i\ldots)}$, I've managed to express the coefficient of $x^3$ as $\frac{1}{2}(i-1)\left(e^{i\pi/24}+e^{5i\pi/24}\right)-\frac{1}{2}(i+1)\left(e^{-i\pi/24}+e^{-5i\pi/24}\right)$ though can't think where to go from there.
A: HINT: If $r$ is a root of a polynomial then $(x-r)$ is a factor.  You have four roots.
A: You may try below :
Say $r_1, r_2, r_3, r_4$ are roots of polynomial,
$P(x) =  \large x^4-\left(\sum r_1\right)x^3 + \left(\sum r_1r_2\right)x^2 -  \left(\sum r_1r_2r_3\right)x +  r_1r_2r_3r_4 $
Note that the roots define the polynomial only upto a constant factor.
A: Observe that $\displaystyle\frac{13\pi}{24}-\frac\pi2=\pi\dfrac{(13-12)}{24}=\frac\pi{24}$
 and $\displaystyle\frac{19\pi}{24}-\frac\pi2=\pi\dfrac{(19-12)}{24}=\frac{7\pi}{24}$
So, $\displaystyle\sin\frac{13\pi}{24}=\sin\left(\frac\pi2+\frac\pi{24}\right)=\cos\frac\pi{24}$ and  $\displaystyle\sin\frac{19\pi}{24}=\sin\left(\frac\pi2+\frac{7\pi}{24}\right)=\cos\frac{7\pi}{24}$
So, we need the four degree equation whose roots are $\displaystyle\sin\frac{7\pi}{24},\cos\frac{7\pi}{24};\sin\frac{\pi}{24},\cos\frac{\pi}{24}$
Now the equation whose roots are $\displaystyle\sin\frac{\pi}{24},\cos\frac{\pi}{24}$ is $$t^2-\left(\sin\frac{\pi}{24}+\cos\frac{\pi}{24}\right)t+\sin\frac{\pi}{24}\cos\frac{\pi}{24}=0$$
Now $\displaystyle\sin\frac{\pi}{24}\cos\frac{\pi}{24}=\frac{\sin\dfrac\pi{12}}2$ and $\displaystyle\dfrac\pi{12}=\frac\pi4-\frac\pi6$
Again, $\displaystyle\sin\frac{\pi}{24}+\cos\frac{\pi}{24}=+\sqrt{1+\sin\dfrac\pi{12}}$
The same method should be applied to find the equation whose roots are $\displaystyle\sin\frac{7\pi}{24},\cos\frac{7\pi}{24}$
