# 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.

• $(x-r_1)(x-r_2)(x-r_3)(x-r_4)$ where the $r$'s are the roots you specified. Commented Aug 21, 2014 at 17:07
• Maybe have a look at this. Commented Aug 21, 2014 at 17:39
• Exactly the kind of thinking shown in your addendum lead me to look at Chebyshev polynomials. The problem is that to get a quartic you must use a quadruple angle. But neither sines nor cosines of the listed angles multiplied by four have equal sines/cosines. Multiplying by six (and hoping to factor out a quadratic) didn't work either. Thus I went with degree 12 (and factoring out a quartic). The numbers you list are the positive roots of that octic, so it may be possible to do something, if you allow square roots in the coefficients. Commented Aug 21, 2014 at 18:14

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.

• I think $R_8(x)=1/2$ works, too? Commented Aug 22, 2014 at 12:47

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.

HINT: If $r$ is a root of a polynomial then $(x-r)$ is a factor. You have four roots.

• I may be wrong but I think the OP wants the coefficients. As far as I know, those $\sin$'s are all rational so $P(x)$ can have integer coefficients. Then again, they may just want it as simple factors :)
– Jam
Commented Aug 21, 2014 at 17:08
• what OP means ?
– idm
Commented Aug 21, 2014 at 17:21
• @kong if Eul Can is correct, find $\sin{\pi \over 24}$ using a half angle formula and then use $\frac{7\pi}{4}=\frac{\pi}{4}+\frac{6\pi}{4}=\frac{\pi}{4}+\frac{3\pi}{2}$ etc. along with the sum of angles formula for the sine function. Commented Aug 21, 2014 at 17:24
• @EulCan The sines are of course not rational, but that does not necesarily prevent from writing the polynomial with integer coefficients. Commented Aug 21, 2014 at 17:29
• @Jean-ClaudeArbaut Sorry, that was stupid of me. I meant algebraic, not rational. Wolfram Alpha gives a representation of them as roots.
– Jam
Commented Aug 21, 2014 at 17:30

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.

• Your summations need different indices. Commented Aug 21, 2014 at 17:36

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 $\displaystyle\sin\frac{7\pi}{24},\cos\frac{7\pi}{24}$

• @Glen_b, The typo has been rectified. Thanks Commented Aug 22, 2014 at 4:45