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Using trignometric identities ( double angle forumlas) one can see that $\cos{2x} = 2 \: \cos^{2}{x} - 1$ can be expressed as a polynomial of $\cos{x}$, where $p(\cos{x})=2 \: \cos^{2}{x}-1$. Then its natural to ask the same question for the sine function.

  • Can we express $\sin{2x}$ as a polynomial of $\sin{x}$?
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Just a small remark: $\cos nx$ is a polynomial in $\cos x$ for any $n$, and so is $\sin nx/\sin x$. These are called Chebyshev polynomials of the first and second kind, respectively, and are usually denoted by $T_n(\cos x)$ and $U_{n-1}(\cos x)$. –  Hans Lundmark Sep 4 '10 at 7:56
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3 Answers

up vote 17 down vote accepted

Just for the sake of completeness, here is a proof that you cannot express it that way.

(For a general proof, see the bottom of this answer).

Suppose we could express $\sin 2x$ as a polynomial in $\sin x$.

i.e.

$\sin x \cos x = P(\sin x)$ for some polynomial $P$.

Since $\sin 0 = 0$, $x$ is a factor of $P(x)$. Hence we get

$\cos x = Q(\sin x)$ for a polynomial $Q$.

Put $x = \frac{\pi}{2}$ and $x=\frac{3\pi}{2}$ we see that $1,-1$ are roots of $Q(x)$.

Thus we see that for 0 < t < 1,

$\sqrt{1-t^2} = (t-1)(t+1)H(t)$ (because $Q(t) = (t-1)(t+1)H(t)$ for some polynomial $H$)

By squaring we see that the degree of the left hand side polynomial is $2$ while the degree of the right hand side polynomial is $>2$. This is not possible, as then their difference, which is not identically zero will have infinite roots.

Thus it is not possible to express $\sin 2x$ as a polynomial in $\sin x$.


General Case:

Here is a proof that in general $ \sin (2nx)$ cannot be written as a polynomial in $ \sin x$ based on the identity provided by J.M:

$$\frac{\sin(2nx)}{2\cos(x)}=(-1)^{\lfloor n-1/2\rfloor}\sum_{j=0}^{\lfloor n-1/2\rfloor}{(-1)^j \sin((2j+1)x)}$$

First notice that $\sin ((2m+1)x)$ can be written as a polynomial in $\sin x$ with integer coefficients (follows easily from the DeMoivre's identity).

Now suppose $ \sin (2nx)$ could be written as a polynomial $P$. Consider the $r^{th}$ derivative of that polynomial at $0$. This must be a rational number, as the $r^{th}$ derivative of $ \sin (2nx)$ at $0$ is rational. Thus the coefficients of $P$ must be rational.

Thus using the above identity we must have that $\cos x$ is of the form $\frac{P(\sin x)}{Q(\sin x)}$ where $P$ and $Q$ are polynomials with rational coefficients. Thus we must have that

$$\sqrt{1-t^2} = \frac{P(t)}{Q(t)}$$

$$ 0 < t < 1$$

Putting $t=\frac{1}{2}$ gives us a rational number on the RHS while an irrational number on the LHS.

Hence it is not possible to represent $ \sin (2nx)$ as a polynomial in $ \sin x$.

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Moron: I did mention that sines of even angle multiples can be expressed as a series of sines of odd multiples, times a cosine. It really all boils down to the question "can the cosine be expressed as a polynomial in sines?" –  J. M. Sep 3 '10 at 14:23
    
@J.M: I don't see immediately how. It boils down to whether cosine can be written as a rational function (P/Q) of sines. How did you get polynomial? But that is an interesting identity you have there! –  Aryabhata Sep 3 '10 at 14:30
    
Hmm, yes, that was too restrictive a question/condition. Nevertheless, your answer cleanly settles the fact that the cosine cannot be a rational function of sines. –  J. M. Sep 3 '10 at 14:49
    
@J.M: I just edited my answer to complete the proof, using your identity. Though I am not sure if my earlier version of the answer proved it. –  Aryabhata Sep 3 '10 at 15:03
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Here's a slightly different way of proving that $\sin nx$ is not a polynomial in $\sin x$ when $n$ is even. Indeed $\sin nx$ cannot be written as $f(\sin x)$ for any function $f$.

To prove this all one has to do is to write down two numbers $a$ and $b$ such that $\sin a=\sin b$ but $\sin na\ne\sin nb$. Let $a=\pi/(2n)$ and $b=\pi-a$. Then $\sin a=\sin b$. But $na=\pi/2$ and $nb=n\pi-\pi/2$ so that $\sin na=1$ and $\sin nb=-1$ (recalling that $n$ is even).

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+1: Nice! I had a feeling there will be something simple like this, but didn't give much thought to it. –  Aryabhata Sep 4 '10 at 17:16
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If $K$ is a commutative ring in which $2\not=0$ and $X$ is an indeterminate, then $X^{2n}-X^{-2n} \notin K[X-X^{-1}]$ for $n>0$. –  Pierre-Yves Gaillard Sep 4 '10 at 18:35
    
Pierre, I think I already know what you're saying, but in a different form, so could you clarify in the context of this question? –  J. M. Sep 5 '10 at 1:58
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Dear J. M.: The question was: "Is $(e^{2niz}-e^{-2niz})/2i$ in $\mathbb C[(e^{iz}-e^{-iz})/2i]$?" This is equivalent to: "Is $X^{2n}-X^{-2n}$ in $\mathbb C[X-X^{-1}]$?" –  Pierre-Yves Gaillard Sep 5 '10 at 4:52
    
Thanks Pierre, I see the correspondence now. –  J. M. Sep 8 '10 at 11:27
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No, you can't get any simpler than $2\sin\;x\;\cos\;x$. In general, only sines of odd angle multiples ($\sin((2k+1)x)$) are expressible as polynomials in $\sin(x)$, because any powers of cosines that may appear are raised to even exponents, and thus the identity $\sin^2(x)=1-\cos^2(x)$ applies, to turn it into an expression entirely made up of powers of sines.

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I leave as an interesting exercise to prove the identity $\frac{\sin(2nx)}{2\cos(x)}=(-1)^{\lfloor n-1/2\rfloor}\sum_{j=0}^{\lfloor n-1/2\rfloor}{(-1)^j \sin((2j+1)x)}$ for $n$ an integer. –  J. M. Sep 3 '10 at 11:07
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