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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.

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$(x-r_1)(x-r_2)(x-r_3)(x-r_4) $ where the $ r $'s are the roots you specified. –  dylan7 Aug 21 '14 at 17:07
Maybe have a look at this. –  jca Aug 21 '14 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. –  Jyrki Lahtonen Aug 21 '14 at 18:14

5 Answers 5

up vote 5 down vote accepted

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.

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I think $R_8(x)=1/2$ works, too? –  Jyrki Lahtonen Aug 22 '14 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







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.

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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.

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Your summations need different indices. –  Paul Sundheim Aug 21 '14 at 17:36

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

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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 :) –  Eul Can Aug 21 '14 at 17:08
what OP means ? –  idm Aug 21 '14 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. –  Paul Sundheim Aug 21 '14 at 17:24
@EulCan The sines are of course not rational, but that does not necesarily prevent from writing the polynomial with integer coefficients. –  jca Aug 21 '14 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. –  Eul Can Aug 21 '14 at 17:30

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

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@Glen_b, The typo has been rectified. Thanks –  lab bhattacharjee Aug 22 '14 at 4:45

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